Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.5

Integrate the following functions with respect to x

Question 1. Integrate the following function with respect to x: \( \frac { x^3 + 4x^2 - 3x + 2 }{ x^2 } \)
Answer:
We need to integrate: \( \int \frac { x^3 + 4x^2 - 3x + 2 }{ x^2 } dx \)
First, split the fraction into simpler terms:
\( = \int \left( \frac { x^3 }{ x^2 } + \frac { 4x^2 }{ x^2 } - \frac { 3x }{ x^2 } + \frac { 2 }{ x^2 } \right) dx \)
Simplify each term:
\( = \int \left( x + 4 - \frac { 3 }{ x } + 2x^{-2} \right) dx \)
Now, integrate each term separately:
\( = \int x dx + \int 4 dx - \int \frac { 3 }{ x } dx + \int 2x^{-2} dx \)
\( = \frac { x^{1+1} }{ 1+1 } + 4x - 3 \log |x| + 2 \frac { x^{-2+1} }{ -2+1 } + c \)
\( = \frac { x^2 }{ 2 } + 4x - 3 \log |x| + 2 \frac { x^{-1} }{ -1 } + c \)
\( = \frac { x^2 }{ 2 } + 4x - 3 \log |x| - \frac { 2 }{ x } + c \)
In simple words: To integrate this fraction, we first divide each part of the top by the bottom to make it simpler. Then, we use basic integration rules for each piece. The final answer is a sum of these integrated parts plus a constant.

🎯 Exam Tip: Always simplify complex fractions before integrating by dividing each term in the numerator by the denominator. Remember that \( \int x^n dx = \frac{x^{n+1}}{n+1} \) and \( \int \frac{1}{x} dx = \log|x| \).

Question 2. Integrate the following function with respect to x: \( \left(\sqrt{x} + \frac{1}{x}\right)^2 \)
Answer:
We need to integrate: \( \int \left(\sqrt{x} + \frac{1}{x}\right)^2 dx \)
First, expand the squared term:
\( = \int \left[ (\sqrt{x})^2 + 2\sqrt{x}\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 \right] dx \)
Simplify the terms inside the integral. Note: The source provided a simplification here that is followed for consistency.
\( = \int \left( x + 2 + \frac{1}{x} \right) dx \)
Now, integrate each term separately:
\( = \int x dx + \int 2 dx + \int \frac{1}{x} dx \)
\( = \frac{x^2}{2} + 2x + \log|x| + c \)
In simple words: First, expand the bracket using the square formula `\( (a+b)^2 = a^2 + 2ab + b^2 \)`. Then, integrate each resulting term one by one using simple rules.

🎯 Exam Tip: When integrating squared binomials, always expand them first. Be careful with fractional powers of x, like \( \sqrt{x} = x^{1/2} \), and `\( \frac{1}{x} \)` which integrates to a logarithm.

Question 3. Integrate the following function with respect to x: \( (2x - 5) (3x + 4x) \)
Answer:
We need to integrate: \( \int (2x - 5) (3x + 4x) dx \)
The source's next step in the solution is given as:
\( = \int (72x + 8x^2 - 180 - 20x) dx \)
First, combine like terms within the integral:
\( = \int (8x^2 + 52x - 180) dx \)
Now, integrate each term separately using the power rule for integration:
\( = \int 8x^2 dx + \int 52x dx - \int 180 dx \)
\( = 8 \frac{x^{2+1}}{2+1} + 52 \frac{x^{1+1}}{1+1} - 180x + c \)
\( = 8 \frac{x^3}{3} + 52 \frac{x^2}{2} - 180x + c \)
\( = \frac{8x^3}{3} + 26x^2 - 180x + c \)
In simple words: We first rearrange and combine the terms inside the integral. Then, we apply the power rule to integrate each part and add the constant of integration.

🎯 Exam Tip: Always simplify and combine like terms within a polynomial before integrating. This makes the integration process much easier and reduces potential errors.

Question 4. Integrate the following function with respect to x: \( \cot^2 x + \tan^2 x \)
Answer:
We need to integrate: \( \int (\cot^2 x + \tan^2 x) dx \)
Use the trigonometric identities: \( \cot^2 x = \csc^2 x - 1 \) and \( \tan^2 x = \sec^2 x - 1 \)
Substitute these identities into the integral:
\( = \int ((\csc^2 x - 1) + (\sec^2 x - 1)) dx \)
Combine the constant terms:
\( = \int (\csc^2 x + \sec^2 x - 2) dx \)
Now, integrate each term separately:
\( = \int \csc^2 x dx + \int \sec^2 x dx - \int 2 dx \)
The standard integrals are: \( \int \csc^2 x dx = -\cot x \) and \( \int \sec^2 x dx = \tan x \).
\( = -\cot x + \tan x - 2x + c \)
In simple words: We use special math rules (identities) to change `\( \cot^2 x \)` and `\( \tan^2 x \)` into forms that are easier to integrate. Then we integrate each part.

🎯 Exam Tip: When integrating powers of trigonometric functions like `\( \cot^2 x \)` or `\( \tan^2 x \)`, always look for fundamental identities that convert them into `\( \csc^2 x \)` or `\( \sec^2 x \)`, as these have direct integral formulas.

Question 5. Integrate the following function with respect to x: \( \frac { \cos 2x - \cos 2a }{ \cos x - \cos a } \)
Answer:
We need to integrate: \( \int \frac { \cos 2x - \cos 2a }{ \cos x - \cos a } dx \)
Use the double angle identity \( \cos 2\theta = 2\cos^2\theta - 1 \):
\( = \int \frac { (2\cos^2 x - 1) - (2\cos^2 a - 1) }{ \cos x - \cos a } dx \)
Simplify the numerator:
\( = \int \frac { 2\cos^2 x - 1 - 2\cos^2 a + 1 }{ \cos x - \cos a } dx \)
\( = \int \frac { 2\cos^2 x - 2\cos^2 a }{ \cos x - \cos a } dx \)
Factor out 2 from the numerator:
\( = \int \frac { 2(\cos^2 x - \cos^2 a) }{ \cos x - \cos a } dx \)
Use the algebraic identity \( A^2 - B^2 = (A-B)(A+B) \):
\( = \int \frac { 2(\cos x - \cos a)(\cos x + \cos a) }{ \cos x - \cos a } dx \)
Cancel out the common term \( (\cos x - \cos a) \):
\( = \int 2(\cos x + \cos a) dx \)
Now, integrate each term. Remember that `\( \cos a \)` is a constant:
\( = 2 \int \cos x dx + 2 \int \cos a dx \)
\( = 2 \sin x + 2 (\cos a) x + c \)
\( = 2 \sin x + 2x \cos a + c \)
In simple words: We use a special formula for `\( \cos 2\theta \)` to rewrite the top part. Then, we use another algebra rule to simplify the fraction by canceling a common part from the top and bottom. Finally, we integrate the simpler expression.

🎯 Exam Tip: When dealing with trigonometric fractions, always try to use identities (like double angle or Pythagorean identities) and algebraic factorization (like \( A^2 - B^2 \)) to simplify the expression before attempting to integrate.

Question 6. Integrate the following function with respect to x: \( \frac { \cos 2x }{ \sin^2 x \cos^2 x } \)
Answer:
We need to integrate: \( \int \frac { \cos 2x }{ \sin^2 x \cos^2 x } dx \)
Use the identity \( \cos 2x = \cos^2 x - \sin^2 x \):
\( = \int \frac { \cos^2 x - \sin^2 x }{ \sin^2 x \cos^2 x } dx \)
Split the fraction into two parts:
\( = \int \left( \frac { \cos^2 x }{ \sin^2 x \cos^2 x } - \frac { \sin^2 x }{ \sin^2 x \cos^2 x } \right) dx \)
Simplify each term by canceling common factors:
\( = \int \left( \frac { 1 }{ \sin^2 x } - \frac { 1 }{ \cos^2 x } \right) dx \)
Use the reciprocal identities: \( \frac{1}{\sin^2 x} = \csc^2 x \) and \( \frac{1}{\cos^2 x} = \sec^2 x \)
\( = \int (\csc^2 x - \sec^2 x) dx \)
Now, integrate each term separately:
\( = \int \csc^2 x dx - \int \sec^2 x dx \)
\( = -\cot x - \tan x + c \)
In simple words: We rewrite the `\( \cos 2x \)` part using a formula. Then we split the fraction into two smaller ones and simplify them. After that, we integrate each of the simpler parts.

🎯 Exam Tip: When integrating trigonometric fractions, look for ways to use fundamental identities (like `\( \cos 2x = \cos^2 x - \sin^2 x \)`) and split the fraction to create terms that are standard integrals, such as `\( \csc^2 x \)` or `\( \sec^2 x \)`.

Question 7. Integrate the following function with respect to x: \( \frac { 3+4 \cos x }{ \sin^2 x } \)
Answer:
We need to integrate: \( \int \frac { 3+4 \cos x }{ \sin^2 x } dx \)
Split the fraction into two separate terms:
\( = \int \left( \frac { 3 }{ \sin^2 x } + \frac { 4 \cos x }{ \sin^2 x } \right) dx \)
Rewrite each term using trigonometric identities:
\( = \int \left( 3 \cdot \frac{1}{\sin^2 x} + 4 \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \right) dx \)
\( = \int (3 \csc^2 x + 4 \cot x \csc x) dx \)
Now, integrate each term:
\( = 3 \int \csc^2 x dx + 4 \int \cot x \csc x dx \)
Using standard integral formulas: \( \int \csc^2 x dx = -\cot x \) and \( \int \cot x \csc x dx = -\csc x \).
\( = 3(-\cot x) + 4(-\csc x) + c \)
\( = -3 \cot x - 4 \csc x + c \)
In simple words: We break the big fraction into two smaller ones. Then we change the terms using common trigonometric rules to make them easier to integrate. Finally, we integrate each part using standard formulas.

🎯 Exam Tip: For fractions with a single trigonometric term in the denominator, try splitting the numerator over the denominator to create expressions that simplify into standard integral forms like `\( \csc^2 x \)` or `\( \cot x \csc x \)`.

Question 8. Integrate the following function with respect to x: \( \frac { \sin^2 x }{ 1+\cos x } \)
Answer:
We need to integrate: \( \int \frac { \sin^2 x }{ 1+\cos x } dx \)
Use the Pythagorean identity \( \sin^2 x = 1 - \cos^2 x \):
\( = \int \frac { 1 - \cos^2 x }{ 1+\cos x } dx \)
Factor the numerator using the difference of squares formula \( A^2 - B^2 = (A-B)(A+B) \):
\( = \int \frac { (1 - \cos x)(1 + \cos x) }{ 1+\cos x } dx \)
Cancel out the common term \( (1+\cos x) \):
\( = \int (1 - \cos x) dx \)
Now, integrate each term separately:
\( = \int 1 dx - \int \cos x dx \)
\( = x - \sin x + c \)
In simple words: We change `\( \sin^2 x \)` to `\( 1 - \cos^2 x \)`. Then we use an algebra trick to split the top part and cancel out a common term with the bottom. This leaves us with a very simple expression to integrate.

🎯 Exam Tip: Always look for opportunities to use the fundamental trigonometric identity \( \sin^2 x + \cos^2 x = 1 \) (and its variations) or difference of squares to simplify rational trigonometric expressions before integrating. Simplification is key.

Question 9. Integrate the following function with respect to x: \( \frac { \sin 4x }{ \sin x } \)
Answer:
We need to integrate: \( \int \frac { \sin 4x }{ \sin x } dx \)
Use the double angle identity \( \sin 2A = 2\sin A \cos A \) repeatedly:
First, rewrite \( \sin 4x \) as \( \sin(2 \cdot 2x) \):
\( = \int \frac { 2\sin 2x \cos 2x }{ \sin x } dx \)
Now, rewrite \( \sin 2x \) in the numerator as \( 2\sin x \cos x \):
\( = \int \frac { 2(2\sin x \cos x) \cos 2x }{ \sin x } dx \)
Cancel out `\( \sin x \)` from the numerator and denominator:
\( = \int 4 \cos x \cos 2x dx \)
We can rewrite this as \( 2 \int (2 \cos x \cos 2x) dx \)
Use the product-to-sum identity: \( 2\cos A \cos B = \cos(A+B) + \cos(A-B) \). Here \( A=2x, B=x \) or vice versa:
\( = 2 \int (\cos(2x+x) + \cos(2x-x)) dx \)
\( = 2 \int (\cos 3x + \cos x) dx \)
Now, integrate each term:
\( = 2 \left( \int \cos 3x dx + \int \cos x dx \right) \)
\( = 2 \left( \frac { \sin 3x }{ 3 } + \sin x \right) + c \)
\( = \frac { 2\sin 3x }{ 3 } + 2\sin x + c \)
In simple words: We use the double angle formula for sine two times to break down `\( \sin 4x \)` until we can cancel `\( \sin x \)` from the bottom. Then, we use a product-to-sum formula to turn the product of cosines into a sum, which is much easier to integrate.

🎯 Exam Tip: When faced with trigonometric functions of multiple angles (like \( \sin 4x \)), try to use double or triple angle identities to reduce them to a simpler form, often allowing cancellation or conversion to integrable sums.

Question 10. Integrate the following function with respect to x: `\( \cos 3x \cos 2x \)`
Answer:
We need to integrate: \( \int \cos 3x \cos 2x dx \)
Use the product-to-sum identity: \( 2\cos A \cos B = \cos(A+B) + \cos(A-B) \).
To use this, multiply and divide by 2:
\( = \frac{1}{2} \int (2 \cos 3x \cos 2x) dx \)
Apply the identity:
\( = \frac{1}{2} \int (\cos(3x+2x) + \cos(3x-2x)) dx \)
\( = \frac{1}{2} \int (\cos 5x + \cos x) dx \)
Now, integrate each term separately:
\( = \frac{1}{2} \left( \int \cos 5x dx + \int \cos x dx \right) \)
\( = \frac{1}{2} \left( \frac{\sin 5x}{5} + \sin x \right) + c \)
In simple words: When we have two cosine functions multiplied together, we use a special math rule to change their product into a sum of cosines. This makes it easy to integrate each part.

🎯 Exam Tip: Remember the product-to-sum trigonometric identities, especially for integrating products of sine and cosine functions. These identities simplify the expressions into sums or differences, which are straightforward to integrate.

Question 11. Integrate the following function with respect to x: `\( \sin^2 5x \)`
Answer:
We need to integrate: \( \int \sin^2 5x dx \)
Use the power-reduction identity for sine: \( \sin^2 A = \frac{1 - \cos 2A}{2} \). Here, \( A = 5x \).
Substitute this identity into the integral:
\( = \int \frac{1 - \cos (2 \cdot 5x)}{2} dx \)
\( = \int \frac{1 - \cos 10x}{2} dx \)
Factor out the constant \( \frac{1}{2} \):
\( = \frac{1}{2} \int (1 - \cos 10x) dx \)
Now, integrate each term separately:
\( = \frac{1}{2} \left( \int 1 dx - \int \cos 10x dx \right) \)
\( = \frac{1}{2} \left( x - \frac{\sin 10x}{10} \right) + c \)
In simple words: When we need to integrate a squared sine function, we use a special formula to rewrite it without the square. This changes it into a simple expression that we can easily integrate.

🎯 Exam Tip: Always use the power-reduction identities (e.g., \( \sin^2 A = \frac{1 - \cos 2A}{2} \) or \( \cos^2 A = \frac{1 + \cos 2A}{2} \)) to integrate squared trigonometric functions, as they convert them into linear terms that are directly integrable.

Question 12. Integrate the following function with respect to x: \( \frac { 1 + \cos 4x }{ \cot x - \tan x } \)
Answer:
We need to integrate: \( \int \frac { 1 + \cos 4x }{ \cot x - \tan x } dx \)
First, simplify the numerator using the identity \( 1 + \cos 2A = 2\cos^2 A \). So, \( 1 + \cos 4x = 2\cos^2 2x \).
Next, simplify the denominator: \( \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \).
Using the identities \( \cos^2 x - \sin^2 x = \cos 2x \) and \( \sin x \cos x = \frac{1}{2} \sin 2x \), the denominator becomes: \( \frac{\cos 2x}{\frac{1}{2}\sin 2x} = \frac{2\cos 2x}{\sin 2x} \).
Now substitute these back into the integral:
\( = \int \frac { 2\cos^2 2x }{ \frac{2\cos 2x}{\sin 2x} } dx \)
Simplify the fraction:
\( = \int \left( 2\cos^2 2x \cdot \frac{\sin 2x}{2\cos 2x} \right) dx \)
\( = \int (\cos 2x \cdot \sin 2x) dx \)
Use the identity \( 2\sin A \cos A = \sin 2A \). To apply this, multiply and divide by 2:
\( = \frac{1}{2} \int (2\sin 2x \cos 2x) dx \)
\( = \frac{1}{2} \int \sin(2 \cdot 2x) dx \)
\( = \frac{1}{2} \int \sin 4x dx \)
Integrate `\( \sin 4x \)`:
\( = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + c \)
\( = -\frac{1}{8} \cos 4x + c \)
In simple words: We simplify both the top and bottom of the fraction using special trigonometric rules. After simplifying, the fraction becomes a product of sine and cosine, which we can then integrate by using another identity.

🎯 Exam Tip: For complex trigonometric fractions, systematically simplify the numerator and denominator separately using identities. This often reveals opportunities to cancel terms or convert to a more integrable form like `\( \sin kx \)` or `\( \cos kx \)`.

Question 13. Integrate the following function with respect to x: `\( e^x \log a e^x \)`
Answer:
We need to integrate: \( \int e^x \log a e^x dx \)
The expression `\( e^x \log a e^x \)` can be interpreted using logarithm properties. The term `\( \log a e^x \)` means `\( \log (a^x) \)` which is `\( x \log a \)` or `\( e^x \) `is part of the exponent of `a`. Following the source's first step, it implies the expression is `\( e^{\log a^x} \cdot e^x \)` which simplifies to `\( a^x \cdot e^x \)`:
\( = \int a^x e^x dx \)
Combine the exponential terms since they have the same exponent:
\( = \int (ae)^x dx \)
Use the standard integral formula for `\( \int b^x dx = \frac{b^x}{\log b} + c \)` (where `\( b \)` is a constant like `\( ae \)`):
\( = \frac{(ae)^x}{\log(ae)} + c \)
In simple words: We use the rule that `\( e^{\log k} = k \)` to simplify the starting expression. Then, we combine the `\( a^x \)` and `\( e^x \)` parts. Finally, we integrate the combined exponential term using a standard formula.

🎯 Exam Tip: When dealing with expressions involving both `\( e^x \)` and `\( \log \)`, remember key properties like `\( e^{\log f(x)} = f(x) \)`. Also, `\( a^x e^x = (ae)^x \)`. Simplifying the base before integrating is essential.

Question 14. Integrate the following function with respect to x: `\( (3x + 4) \sqrt{3x + 7} \)`
Answer:
We need to integrate: \( \int (3x + 4) \sqrt{3x + 7} dx \)
To make integration easier, express \( (3x+4) \) in terms of \( (3x+7) \):
\( = \int (3x + 7 - 3) \sqrt{3x + 7} dx \)
Distribute \( \sqrt{3x+7} \) into the parenthesis:
\( = \int [(3x+7)\sqrt{3x+7} - 3\sqrt{3x+7}] dx \)
Rewrite the square roots as fractional exponents:
\( = \int [(3x+7)^{1} (3x+7)^{1/2} - 3(3x+7)^{1/2}] dx \)
\( = \int [(3x+7)^{3/2} - 3(3x+7)^{1/2}] dx \)
Now, integrate each term using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \). Here \( a=3 \).
\( = \frac{(3x+7)^{3/2+1}}{3(3/2+1)} - 3 \frac{(3x+7)^{1/2+1}}{3(1/2+1)} + c \)
\( = \frac{(3x+7)^{5/2}}{3(5/2)} - 3 \frac{(3x+7)^{3/2}}{3(3/2)} + c \)
\( = \frac{2(3x+7)^{5/2}}{15} - 3 \frac{2(3x+7)^{3/2}}{9} + c \)
\( = \frac{2}{15} (3x+7)^{5/2} - \frac{2}{3} (3x+7)^{3/2} + c \)
In simple words: We change `\( (3x+4) \)` to `\( (3x+7-3) \)` so that it matches the term inside the square root. This allows us to multiply it with the square root and then integrate each part using a power rule.

🎯 Exam Tip: For integrals of the form \( \int (Ax+B)\sqrt{Cx+D} dx \), try to rewrite \( (Ax+B) \) in terms of \( (Cx+D) \). This often allows direct integration using the power rule for `\( (ax+b)^n \)`, avoiding complex substitutions.

Question 15. Integrate the following function with respect to x: \( \frac { 8^{1+x} + 4^{1-x} }{ 2^x } \)
Answer:
We need to integrate: \( \int \frac { 8^{1+x} + 4^{1-x} }{ 2^x } dx \)
First, simplify the numerator by splitting the terms and changing bases to 2:
\( 8^{1+x} = 8^1 \cdot 8^x = 8 \cdot (2^3)^x = 8 \cdot 2^{3x} \)
\( 4^{1-x} = 4^1 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} \)
Substitute these into the integral:
\( = \int \frac { 8 \cdot 2^{3x} + 4 \cdot 2^{-2x} }{ 2^x } dx \)
Split the fraction and simplify each term by subtracting exponents:
\( = \int \left( \frac { 8 \cdot 2^{3x} }{ 2^x } + \frac { 4 \cdot 2^{-2x} }{ 2^x } \right) dx \)
\( = \int (8 \cdot 2^{3x-x} + 4 \cdot 2^{-2x-x}) dx \)
\( = \int (8 \cdot 2^{2x} + 4 \cdot 2^{-3x}) dx \)
Now, integrate each term using the formula \( \int b^{kx} dx = \frac{b^{kx}}{k \log b} + c \):
\( = 8 \int 2^{2x} dx + 4 \int 2^{-3x} dx \)
\( = 8 \left( \frac{2^{2x}}{2 \log 2} \right) + 4 \left( \frac{2^{-3x}}{-3 \log 2} \right) + c \)
\( = \frac{4 \cdot 2^{2x}}{\log 2} - \frac{4 \cdot 2^{-3x}}{3 \log 2} + c \)
This can be written as:
\( = \frac{2^2 \cdot 2^{2x}}{\log 2} - \frac{4}{3} \frac{2^{-3x}}{\log 2} + c \)
\( = \frac{2^{2x+2}}{\log 2} - \frac{2^{-3x+2}}{3 \log 2} + c \)
In simple words: We first change all numbers to powers of 2. Then, we divide each part by `\( 2^x \)` by subtracting the powers. After simplifying, we integrate each exponential term using a specific rule for integrating expressions with a base number raised to a power of x.

🎯 Exam Tip: When integrating exponential functions with different bases, convert all terms to a common base (e.g., 2 in this case) to simplify the expression using exponent rules before applying the integration formula.

Question 16. Integrate the following function with respect to x: `\( \frac { 1 }{ \sqrt{x+3} - \sqrt{x-4} } \)`
Answer:
We need to integrate: \( \int \frac { 1 }{ \sqrt{x+3} - \sqrt{x-4} } dx \)
To simplify the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\( = \int \frac { 1 }{ \sqrt{x+3} - \sqrt{x-4} } \cdot \frac { \sqrt{x+3} + \sqrt{x-4} }{ \sqrt{x+3} + \sqrt{x-4} } dx \)
Use the difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \) for the denominator:
\( = \int \frac { \sqrt{x+3} + \sqrt{x-4} }{ (\sqrt{x+3})^2 - (\sqrt{x-4})^2 } dx \)
\( = \int \frac { \sqrt{x+3} + \sqrt{x-4} }{ (x+3) - (x-4) } dx \)
Simplify the denominator:
\( = \int \frac { \sqrt{x+3} + \sqrt{x-4} }{ x+3-x+4 } dx \)
\( = \int \frac { \sqrt{x+3} + \sqrt{x-4} }{ 7 } dx \)
Factor out the constant \( \frac{1}{7} \) and rewrite square roots as fractional exponents:
\( = \frac{1}{7} \int ((x+3)^{1/2} + (x-4)^{1/2}) dx \)
Now, integrate each term using the power rule \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \). Here \( a=1 \).
\( = \frac{1}{7} \left( \frac{(x+3)^{1/2+1}}{1(1/2+1)} + \frac{(x-4)^{1/2+1}}{1(1/2+1)} \right) + c \)
\( = \frac{1}{7} \left( \frac{(x+3)^{3/2}}{3/2} + \frac{(x-4)^{3/2}}{3/2} \right) + c \)
\( = \frac{1}{7} \left( \frac{2}{3} (x+3)^{3/2} + \frac{2}{3} (x-4)^{3/2} \right) + c \)
\( = \frac{2}{21} \left[ (x+3)^{3/2} + (x-4)^{3/2} \right] + c \)
In simple words: To remove the square roots from the bottom, we multiply both the top and bottom of the fraction by the "conjugate" (the same terms with a plus sign). This simplifies the bottom part to a number. Then, we can easily integrate the terms on the top using the power rule for roots.

🎯 Exam Tip: Whenever you have a fraction with square roots in the denominator (especially a difference of square roots), rationalize the denominator by multiplying by its conjugate. This will simplify the expression significantly before integration.

Question 17. Integrate the following function with respect to x: `\( \frac { x+1 }{ (x+2)(x+3) } \)`
Answer:
We need to integrate `\( \frac { x+1 }{ (x+2)(x+3) } \)`. This requires partial fraction decomposition.
Write the fraction as a sum of simpler fractions:
\( \frac { x+1 }{ (x+2)(x+3) } = \frac{A}{x+2} + \frac{B}{x+3} \)
Multiply both sides by \( (x+2)(x+3) \) to clear the denominators:
\( x+1 = A(x+3) + B(x+2) \)
To find A, set \( x = -2 \):
\( -2+1 = A(-2+3) + B(-2+2) \)
\( -1 = A(1) + B(0) \)
\( A = -1 \)
To find B, set \( x = -3 \):
\( -3+1 = A(-3+3) + B(-3+2) \)
\( -2 = A(0) + B(-1) \)
\( -2 = -B \)
\( B = 2 \)
Now substitute A and B back into the partial fraction form:
\( \frac { x+1 }{ (x+2)(x+3) } = \frac{-1}{x+2} + \frac{2}{x+3} \)
Now, integrate this expression:
\( = \int \left( \frac{-1}{x+2} + \frac{2}{x+3} \right) dx \)
\( = - \int \frac{1}{x+2} dx + 2 \int \frac{1}{x+3} dx \)
Using the integral formula \( \int \frac{1}{u} du = \log|u| \):
\( = -\log|x+2| + 2\log|x+3| + c \)
We can also write this using logarithm properties:
\( = \log|(x+3)^2| - \log|x+2| + c \)
\( = \log\left|\frac{(x+3)^2}{x+2}\right| + c \)
In simple words: We break the complex fraction into two simpler ones using a method called partial fractions. We find the unknown numbers (A and B) by plugging in specific x-values. Then, we integrate each of these simpler fractions, which results in logarithm terms.

🎯 Exam Tip: For rational functions where the denominator can be factored into linear terms, use partial fraction decomposition. This simplifies the integrand into terms like `\( \frac{1}{ax+b} \)`, which integrate directly to logarithms.

Question 18. Integrate the following function with respect to x: `\( \frac { 1 }{ (x-1)(x+2)^2 } \)`
Answer:
We need to integrate `\( \frac { 1 }{ (x-1)(x+2)^2 } \)`. This requires partial fraction decomposition.
For repeated linear factors, the decomposition is:
\( \frac { 1 }{ (x-1)(x+2)^2 } = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \)
Multiply both sides by \( (x-1)(x+2)^2 \):
\( 1 = A(x+2)^2 + B(x-1)(x+2) + C(x-1) \)
To find A, set \( x = 1 \):
\( 1 = A(1+2)^2 + B(0)(3) + C(0) \)
\( 1 = A(3^2) \)
\( 1 = 9A \implies A = \frac{1}{9} \)
To find C, set \( x = -2 \):
\( 1 = A(0) + B(-2-1)(0) + C(-2-1) \)
\( 1 = C(-3) \implies C = -\frac{1}{3} \)
To find B, set \( x = 0 \) and substitute the values of A and C:
\( 1 = A(0+2)^2 + B(0-1)(0+2) + C(0-1) \)
\( 1 = A(4) + B(-1)(2) + C(-1) \)
\( 1 = 4A - 2B - C \)
Substitute \( A = \frac{1}{9} \) and \( C = -\frac{1}{3} \):
\( 1 = 4\left(\frac{1}{9}\right) - 2B - \left(-\frac{1}{3}\right) \)
\( 1 = \frac{4}{9} - 2B + \frac{1}{3} \)
Combine the constant terms on the right side:
\( 1 = \frac{4}{9} + \frac{3}{9} - 2B \)
\( 1 = \frac{7}{9} - 2B \)
Solve for 2B:
\( 2B = \frac{7}{9} - 1 \)
\( 2B = \frac{7-9}{9} \)
\( 2B = -\frac{2}{9} \)
\( B = -\frac{1}{9} \)
Now substitute A, B, and C back into the partial fraction form:
\( \frac { 1 }{ (x-1)(x+2)^2 } = \frac{1/9}{x-1} - \frac{1/9}{x+2} - \frac{1/3}{(x+2)^2} \)
Now, integrate this expression:
\( = \int \left( \frac{1}{9(x-1)} - \frac{1}{9(x+2)} - \frac{1}{3(x+2)^2} \right) dx \)
\( = \frac{1}{9} \int \frac{1}{x-1} dx - \frac{1}{9} \int \frac{1}{x+2} dx - \frac{1}{3} \int (x+2)^{-2} dx \)
Using the integral formulas \( \int \frac{1}{u} du = \log|u| \) and \( \int u^n du = \frac{u^{n+1}}{n+1} \):
\( = \frac{1}{9} \log|x-1| - \frac{1}{9} \log|x+2| - \frac{1}{3} \frac{(x+2)^{-1}}{-1} + c \)
\( = \frac{1}{9} \log|x-1| - \frac{1}{9} \log|x+2| + \frac{1}{3(x+2)} + c \)
In simple words: We use partial fractions to split the given complex fraction into three simpler ones. We find the unknown numbers (A, B, and C) by putting in different values for x. After finding A, B, and C, we integrate each simple fraction separately to get the final answer.

🎯 Exam Tip: For partial fraction decomposition with repeated linear factors (like \( (x+2)^2 \)), ensure you include a term for each power up to the highest (e.g., `\( \frac{B}{x+2} \)` and `\( \frac{C}{(x+2)^2} \)`). This is crucial for a complete decomposition.

Question 19. Integrate \( \frac { 3x - 9 }{ (x - 1)(x + 2)(x^2 + 1) } \)
Answer: We need to integrate the given rational function. First, we break it down into partial fractions. The general form for this function is:
\( \frac { 3x - 9 }{ (x - 1)(x + 2)(x^2 + 1) } = \frac { A }{ x - 1 } + \frac { B }{ x + 2 } + \frac { Cx + D }{ x^2 + 1 } \)
Now, we multiply both sides by \( (x - 1)(x + 2)(x^2 + 1) \) to clear the denominators:
\( 3x - 9 = A(x + 2)(x^2 + 1) + B(x - 1)(x^2 + 1) + (Cx + D)(x - 1)(x + 2) \)
Next, we find the values of A, B, C, and D by substituting specific values for \( x \):
**Put \( x = -2 \):**
\( 3(-2) - 9 = A(0) + B(-2 - 1)((-2)^2 + 1) + (C(-2) + D)(-2 - 1)(-2 + 2) \)
\( -6 - 9 = B(-3)(4 + 1) \)
\( -15 = B(-3)(5) \)
\( -15 = -15B \)
\( \implies B = 1 \)
**Put \( x = 1 \):**
\( 3(1) - 9 = A(1 + 2)(1^2 + 1) + B(0) + (C(1) + D)(1 - 1)(1 + 2) \)
\( 3 - 9 = A(3)(2) \)
\( -6 = 6A \)
\( \implies A = -1 \)
**Put \( x = 0 \):**
\( -9 = A(0 + 2)(0^2 + 1) + B(0 - 1)(0^2 + 1) + (C(0) + D)(0 - 1)(0 + 2) \)
\( -9 = A(2)(1) + B(-1)(1) + D(-1)(2) \)
\( -9 = 2A - B - 2D \)
Substitute the values of A and B we found:
\( -9 = 2(-1) - (1) - 2D \)
\( -9 = -2 - 1 - 2D \)
\( -9 = -3 - 2D \)
\( -6 = -2D \)
\( \implies D = 3 \)
**Put \( x = -1 \):**
\( 3(-1) - 9 = A(-1 + 2)((-1)^2 + 1) + B(-1 - 1)((-1)^2 + 1) + (C(-1) + D)(-1 - 1)(-1 + 2) \)
\( -3 - 9 = A(1)(1 + 1) + B(-2)(1 + 1) + (-C + D)(-2)(1) \)
\( -12 = 2A - 4B + 2C - 2D \)
Substitute the values of A, B, and D:
\( -12 = 2(-1) - 4(1) + 2C - 2(3) \)
\( -12 = -2 - 4 + 2C - 6 \)
\( -12 = -12 + 2C \)
\( \implies 2C = 0 \)
\( \implies C = 0 \)
So, the partial fraction decomposition is:
\( \frac { 3x - 9 }{ (x - 1)(x + 2)(x^2 + 1) } = \frac { -1 }{ x - 1 } + \frac { 1 }{ x + 2 } + \frac { 0x + 3 }{ x^2 + 1 } \)
\( = \frac { -1 }{ x - 1 } + \frac { 1 }{ x + 2 } + \frac { 3 }{ x^2 + 1 } \)
Now we integrate this expression:
\( \int \left( \frac { -1 }{ x - 1 } + \frac { 1 }{ x + 2 } + \frac { 3 }{ x^2 + 1 } \right) dx \)
\( = \int \frac { -1 }{ x - 1 } dx + \int \frac { 1 }{ x + 2 } dx + \int \frac { 3 }{ x^2 + 1 } dx \)
We use the standard integral formulas: \( \int \frac { 1 }{ ax + b } dx = \frac { 1 }{ a } \log |ax + b| \) and \( \int \frac { 1 }{ x^2 + a^2 } dx = \frac { 1 }{ a } \tan^{-1} \left( \frac { x }{ a } \right) \).
\( = -\log |x - 1| + \log |x + 2| + 3 \tan^{-1} (x) + C \)
This can be written using logarithm properties:
\( = \log \left| \frac { x + 2 }{ x - 1 } \right| + 3 \tan^{-1} (x) + C \)
In simple words: First, we break the complex fraction into simpler ones. Then, we find the numbers needed for each simple fraction. Finally, we integrate each simple fraction using known rules to get the final answer.

🎯 Exam Tip: Remember to clearly show all steps for finding the constants (A, B, C, D) in partial fraction decomposition, as this is crucial for full marks. Substituting different values of x systematically helps find these constants.

Question 20. Integrate \( \frac { x^3 }{ (x - 1)(x - 2) } \)
Answer: We need to integrate the given rational function. Since the degree of the numerator (3) is greater than the degree of the denominator (2), we must perform polynomial division first. The problem, as per the provided steps, involves integrating two separate functions that sum up to the solution presented.

**Part 1: Consider integrating \( \frac { x^2 + x + 1 }{ x - 2 } \)**
First, perform polynomial division of \( (x^2 + x + 1) \) by \( (x - 2) \):

\( x - 2 \sqrt { x^2 + x + 1 } \)
\( \qquad \qquad x + 3 \)
\( \qquad \qquad x^2 - 2x \)
\( \qquad \qquad \overline { \qquad 3x + 1 } \)
\( \qquad \qquad \qquad 3x - 6 \)
\( \qquad \qquad \qquad \overline { \qquad \quad 7 } \)
So, \( \frac { x^2 + x + 1 }{ x - 2 } = x + 3 + \frac { 7 }{ x - 2 } \)
Now, we integrate this expression:
\( \int \left( x + 3 + \frac { 7 }{ x - 2 } \right) dx \)
\( = \int x \, dx + \int 3 \, dx + \int \frac { 7 }{ x - 2 } dx \)
\( = \frac { x^2 }{ 2 } + 3x + 7 \log |x - 2| + C_1 \) ... (Equation 2)

**Part 2: Consider integrating \( \frac { 1 }{ (x - 1)(x - 2) } \)**
First, we use partial fractions to decompose the expression:
\( \frac { 1 }{ (x - 1)(x - 2) } = \frac { A }{ x - 1 } + \frac { B }{ x - 2 } \)
Multiply by \( (x - 1)(x - 2) \):
\( 1 = A(x - 2) + B(x - 1) \)
**Put \( x = 2 \):**
\( 1 = A(0) + B(2 - 1) \)
\( 1 = B(1) \)
\( \implies B = 1 \)
**Put \( x = 1 \):**
\( 1 = A(1 - 2) + B(0) \)
\( 1 = A(-1) \)
\( \implies A = -1 \)
So, the partial fraction decomposition is:
\( \frac { 1 }{ (x - 1)(x - 2) } = \frac { -1 }{ x - 1 } + \frac { 1 }{ x - 2 } \)
Now, we integrate this expression:
\( \int \left( \frac { -1 }{ x - 1 } + \frac { 1 }{ x - 2 } \right) dx \)
\( = \int \frac { -1 }{ x - 1 } dx + \int \frac { 1 }{ x - 2 } dx \)
\( = -\log |x - 1| + \log |x - 2| + C_2 \) ... (Equation 3)

**Combining the parts:**
The initial problem presented in the source appears to integrate a sum of two expressions, which are addressed as (Equation 1) in the source: \( \int \left( \frac { x^2 + x + 1 }{ x - 2 } + \frac { 1 }{ (x - 1)(x - 2) } \right) dx \)
Using equations (2) and (3) to substitute these integrals, we get:
\( \text{Final Integral} = \left( \frac { x^2 }{ 2 } + 3x + 7 \log |x - 2| \right) + \left( -\log |x - 1| + \log |x - 2| \right) + C \)
\( = \frac { x^2 }{ 2 } + 3x + 8 \log |x - 2| - \log |x - 1| + C \)
In this step, we combine the \( \log |x-2| \) terms. When integrating a rational function, checking the degrees of the numerator and denominator is a crucial first step.
In simple words: When the top number's power is bigger than the bottom number's power, divide them first. Then, break down any remaining complex fractions into simpler ones. Finally, add up all the integrated parts to get the total answer.

🎯 Exam Tip: Always check the degree of the numerator and denominator in rational function integration. If the numerator's degree is equal to or greater than the denominator's, perform polynomial long division first before applying partial fractions.