Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.7

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7

Question 1. Integrate the following with respect to x. \( \frac { 1 }{9-16x^2} \)
Answer:
To integrate \( \frac { 1 }{9-16x^2} \) with respect to x, we first rewrite the denominator to match a standard integral form. We can factor out 16 from the denominator.
\[ \int \frac { 1 }{9-16x^2} dx \] Now, take out 16 from the denominator: \[ = \int \frac { 1 }{16 \left( \frac{9}{16} - x^2 \right)} dx \] \[ = \frac { 1 }{16} \int \frac { 1 }{\left( \frac{3}{4} \right)^2 - x^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{a^2 - x^2} dx = \frac { 1 }{2a} \log \left| \frac { a+x }{ a-x } \right| + c \).
Here, \( a = \frac{3}{4} \).
\[ = \frac { 1 }{16} \left[ \frac { 1 }{2 \left( \frac{3}{4} \right)} \log \left| \frac { \frac{3}{4}+x }{ \frac{3}{4}-x } \right| \right] + c \] Next, simplify the term outside the logarithm:
\[ = \frac { 1 }{16} \left[ \frac { 1 }{\frac{3}{2}} \log \left| \frac { \frac{3+4x}{4} }{ \frac{3-4x}{4} } \right| \right] + c \] \[ = \frac { 1 }{16} \left[ \frac { 2 }{3} \log \left| \frac { 3+4x }{ 3-4x } \right| \right] + c \] Finally, multiply the constants:
\[ = \frac { 1 }{24} \log \left| \frac { 3+4x }{ 3-4x } \right| + c \] The result is a logarithmic expression. This type of integral is very common in calculus.
In simple words: To solve this, we first change the bottom part of the fraction to fit a known formula. Then, we use that formula to find the integral, which involves a logarithm.

🎯 Exam Tip: Remember to factor out any coefficients of \(x^2\) in the denominator to simplify the expression into a standard form before applying integral formulas. This step is often missed.

Question 2. \( \frac { 1} {9-8x-x^2} \)
Answer:
To integrate \( \frac { 1} {9-8x-x^2} \) with respect to x, we complete the square in the denominator. This helps us transform the expression into a standard integration form.
First, consider the denominator: \( 9-8x-x^2 \).
We can rewrite this by factoring out a negative sign from the \(x\) terms:
\( 9-8x-x^2 = 9 - (x^2 + 8x) \)
To complete the square for \( x^2 + 8x \), we add and subtract \( \left( \frac{8}{2} \right)^2 = 4^2 = 16 \):
\( x^2 + 8x = x^2 + 8x + 16 - 16 = (x+4)^2 - 16 \)
Now substitute this back into the denominator:
\( 9 - (x^2 + 8x) = 9 - [(x+4)^2 - 16] \)
\( = 9 - (x+4)^2 + 16 \)
\( = 25 - (x+4)^2 \)
This can be written as \( 5^2 - (x+4)^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{5^2-(x+4)^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{a^2 - x^2} dx = \frac { 1 }{2a} \log \left| \frac { a+x }{ a-x } \right| + c \).
Here, \( a = 5 \) and \( x \) is replaced by \( (x+4) \).
\[ = \frac { 1 }{2(5)} \log \left| \frac { 5+(x+4) }{ 5-(x+4) } \right| + c \] \[ = \frac { 1 }{10} \log \left| \frac { 5+x+4 }{ 5-x-4 } \right| + c \] \[ = \frac { 1 }{10} \log \left| \frac { 9+x }{ 1-x } \right| + c \] Completing the square is a key technique for solving these types of integrals.
In simple words: First, we change the bottom part of the fraction by adding and subtracting a number to make it a perfect square. This helps us fit it into a standard integration formula, which then gives us a logarithm as the answer.

🎯 Exam Tip: Always be careful with the negative signs when completing the square, especially when \(x^2\) has a negative coefficient. A small error in signs can lead to a completely wrong answer.

Question 3. \( \frac { 1 }{2x^2-9} \)
Answer:
To integrate \( \frac { 1 }{2x^2-9} \) with respect to x, we first adjust the denominator to fit a standard integral formula. We factor out the coefficient of \(x^2\).
\[ \int \frac { 1 }{2x^2-9} dx \] Factor out 2 from the denominator:
\[ = \int \frac { 1 }{2 \left( x^2 - \frac{9}{2} \right)} dx \] \[ = \frac { 1 }{2} \int \frac { 1 }{x^2 - \left( \frac{3}{\sqrt{2}} \right)^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{x^2 - a^2} dx = \frac { 1 }{2a} \log \left| \frac { x-a }{ x+a } \right| + c \).
Here, \( a = \frac{3}{\sqrt{2}} \).
\[ = \frac { 1 }{2} \left[ \frac { 1 }{2 \left( \frac{3}{\sqrt{2}} \right)} \log \left| \frac { x - \frac{3}{\sqrt{2}} }{ x + \frac{3}{\sqrt{2}} } \right| \right] + c \] Next, simplify the constant term outside the logarithm:
\[ = \frac { 1 }{2} \left[ \frac { \sqrt{2} }{6} \log \left| \frac { \frac{\sqrt{2}x-3}{\sqrt{2}} }{ \frac{\sqrt{2}x+3}{\sqrt{2}} } \right| \right] + c \] \[ = \frac { \sqrt{2} }{12} \log \left| \frac { \sqrt{2}x-3 }{ \sqrt{2}x+3 } \right| + c \] To further simplify, we can multiply the numerator and denominator by \( \frac{\sqrt{2}}{\sqrt{2}} \) to rationalize the denominator of the constant term outside:
\[ = \frac { \sqrt{2} }{12} = \frac { \sqrt{2} \times \sqrt{2} }{12 \times \sqrt{2}} = \frac { 2 }{12\sqrt{2}} = \frac { 1 }{6\sqrt{2}} \] So, the final answer is:
\[ = \frac { 1 }{6\sqrt{2}} \log \left| \frac { \sqrt{2}x-3 }{ \sqrt{2}x+3 } \right| + c \] This demonstrates how factoring out constants simplifies complex integrals.
In simple words: We first take out the number 2 from the bottom of the fraction. This makes the fraction look like a standard formula for integration. After using the formula, we simplify the numbers to get the final answer, which is a logarithm.

🎯 Exam Tip: Always remember to rationalize the denominator of constants in your final answer if they involve square roots, as it is considered good mathematical practice.

Question 4. \( \frac { 1 }{x^2-x-2} \)
Answer:
To integrate \( \frac { 1 }{x^2-x-2} \) with respect to x, we complete the square in the denominator. This process converts the expression into a standard form for integration.
First, consider the denominator: \( x^2-x-2 \).
To complete the square for \( x^2-x \), we add and subtract \( \left( \frac{-1}{2} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \):
\( x^2-x-2 = x^2-x+\frac{1}{4}-\frac{1}{4}-2 \)
\( = \left( x-\frac{1}{2} \right)^2 - \frac{1}{4} - \frac{8}{4} \)
\( = \left( x-\frac{1}{2} \right)^2 - \frac{9}{4} \)
This can be written as \( \left( x-\frac{1}{2} \right)^2 - \left( \frac{3}{2} \right)^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{\left( x-\frac{1}{2} \right)^2 - \left( \frac{3}{2} \right)^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{x^2 - a^2} dx = \frac { 1 }{2a} \log \left| \frac { x-a }{ x+a } \right| + c \).
Here, \( x \) is replaced by \( \left( x-\frac{1}{2} \right) \) and \( a = \frac{3}{2} \).
\[ = \frac { 1 }{2 \left( \frac{3}{2} \right)} \log \left| \frac { \left( x-\frac{1}{2} \right) - \frac{3}{2} }{ \left( x-\frac{1}{2} \right) + \frac{3}{2} } \right| + c \] \[ = \frac { 1 }{3} \log \left| \frac { x - \frac{1+3}{2} }{ x + \frac{-1+3}{2} } \right| + c \] \[ = \frac { 1 }{3} \log \left| \frac { x - \frac{4}{2} }{ x + \frac{2}{2} } \right| + c \] \[ = \frac { 1 }{3} \log \left| \frac { x-2 }{ x+1 } \right| + c \] This illustrates the power of completing the square in simplifying integrals.
In simple words: We change the bottom part of the fraction to a perfect square minus another square. Then, we use a special integration rule that gives us a logarithm as the answer, making sure to combine the numbers correctly.

🎯 Exam Tip: When completing the square, ensure the coefficient of \(x^2\) is 1. If not, factor it out first. Also, be careful with adding and subtracting the square of half the coefficient of x.

Question 5. \( \frac { 1 }{x^2+3x+2} \)
Answer:
To integrate \( \frac { 1 }{x^2+3x+2} \) with respect to x, we first complete the square in the denominator. This will help us use a standard integration formula.
Consider the denominator: \( x^2+3x+2 \).
To complete the square for \( x^2+3x \), we add and subtract \( \left( \frac{3}{2} \right)^2 = \frac{9}{4} \):
\( x^2+3x+2 = x^2+3x+\frac{9}{4}-\frac{9}{4}+2 \)
\( = \left( x+\frac{3}{2} \right)^2 - \frac{9}{4} + \frac{8}{4} \)
\( = \left( x+\frac{3}{2} \right)^2 - \frac{1}{4} \)
This can be written as \( \left( x+\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{\left( x+\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{x^2 - a^2} dx = \frac { 1 }{2a} \log \left| \frac { x-a }{ x+a } \right| + c \).
Here, \( x \) is replaced by \( \left( x+\frac{3}{2} \right) \) and \( a = \frac{1}{2} \).
\[ = \frac { 1 }{2 \left( \frac{1}{2} \right)} \log \left| \frac { \left( x+\frac{3}{2} \right) - \frac{1}{2} }{ \left( x+\frac{3}{2} \right) + \frac{1}{2} } \right| + c \] \[ = \frac { 1 }{1} \log \left| \frac { x + \frac{3-1}{2} }{ x + \frac{3+1}{2} } \right| + c \] \[ = \log \left| \frac { x + \frac{2}{2} }{ x + \frac{4}{2} } \right| + c \] \[ = \log \left| \frac { x+1 }{ x+2 } \right| + c \] This method helps convert complex denominators into simpler forms for integration.
In simple words: We change the bottom part of the fraction by making it a perfect square. This allows us to use a special math rule for integrals that results in a logarithm.

🎯 Exam Tip: When completing the square, remember that \( (a+b)^2 = a^2+2ab+b^2 \). Identify the '2ab' term correctly to find 'b' (half the coefficient of x) for completing the square.

Question 6. \( \frac { 1 }{2x^2+6x-8} \)
Answer:
To integrate \( \frac { 1 }{2x^2+6x-8} \) with respect to x, we start by simplifying the denominator through factoring and then completing the square.
First, consider the denominator: \( 2x^2+6x-8 \).
Factor out 2 from the denominator:
\( 2x^2+6x-8 = 2(x^2+3x-4) \)
Now, complete the square for the term inside the parenthesis, \( x^2+3x-4 \).
To complete the square for \( x^2+3x \), we add and subtract \( \left( \frac{3}{2} \right)^2 = \frac{9}{4} \):
\( x^2+3x-4 = x^2+3x+\frac{9}{4}-\frac{9}{4}-4 \)
\( = \left( x+\frac{3}{2} \right)^2 - \frac{9}{4} - \frac{16}{4} \)
\( = \left( x+\frac{3}{2} \right)^2 - \frac{25}{4} \)
This can be written as \( \left( x+\frac{3}{2} \right)^2 - \left( \frac{5}{2} \right)^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{2 \left[ \left( x+\frac{3}{2} \right)^2 - \left( \frac{5}{2} \right)^2 \right]} dx \] \[ = \frac { 1 }{2} \int \frac { 1 }{\left( x+\frac{3}{2} \right)^2 - \left( \frac{5}{2} \right)^2} dx \] We use the standard integral formula: \( \int \frac { 1 }{x^2 - a^2} dx = \frac { 1 }{2a} \log \left| \frac { x-a }{ x+a } \right| + c \).
Here, \( x \) is replaced by \( \left( x+\frac{3}{2} \right) \) and \( a = \frac{5}{2} \).
\[ = \frac { 1 }{2} \left[ \frac { 1 }{2 \left( \frac{5}{2} \right)} \log \left| \frac { \left( x+\frac{3}{2} \right) - \frac{5}{2} }{ \left( x+\frac{3}{2} \right) + \frac{5}{2} } \right| \right] + c \] \[ = \frac { 1 }{2} \left[ \frac { 1 }{5} \log \left| \frac { x + \frac{3-5}{2} }{ x + \frac{3+5}{2} } \right| \right] + c \] \[ = \frac { 1 }{10} \log \left| \frac { x + \frac{-2}{2} }{ x + \frac{8}{2} } \right| + c \] \[ = \frac { 1 }{10} \log \left| \frac { x-1 }{ x+4 } \right| + c \] This shows how factoring and completing the square work together to solve integrals.
In simple words: First, we pull out the number 2 from the bottom of the fraction. Then, we make the remaining part into a perfect square minus another square. Using a special integral rule, we find the answer, which is a logarithm.

🎯 Exam Tip: Always factor out any common coefficients from the \(x^2\) term before attempting to complete the square, as this simplifies the process and avoids errors with fractions.

Question 7. \( \frac { e^x }{e^{2x}-9} \)
Answer:
To integrate \( \frac { e^x }{e^{2x}-9} \) with respect to x, we use a substitution method. This transforms the integral into a simpler form that matches a standard formula.
Let \( t = e^x \).
Then, \( dt = e^x dx \).
Also, \( e^{2x} = (e^x)^2 = t^2 \).
Substitute these into the integral:
\[ \int \frac { e^x }{e^{2x}-9} dx = \int \frac { dt }{t^2-3^2} \] We use the standard integral formula: \( \int \frac { 1 }{x^2 - a^2} dx = \frac { 1 }{2a} \log \left| \frac { x-a }{ x+a } \right| + c \).
Here, \( t \) is the variable and \( a=3 \).
\[ = \frac { 1 }{2(3)} \log \left| \frac { t-3 }{ t+3 } \right| + c \] \[ = \frac { 1 }{6} \log \left| \frac { t-3 }{ t+3 } \right| + c \] Now, substitute back \( t = e^x \):
\[ = \frac { 1 }{6} \log \left| \frac { e^x-3 }{ e^x+3 } \right| + c \] The substitution method is very effective for integrals involving exponential functions.
In simple words: We change the variable from \(x\) to \(t = e^x\), which makes the integral look like a simple fraction with \(t\). Then, we use a standard rule to solve it and finally change \(t\) back to \(e^x\) for the answer.

🎯 Exam Tip: For integrals involving \(e^x\) in both the numerator and denominator, a common strategy is to use the substitution \(t = e^x\) to simplify the expression into a more manageable algebraic form.

Question 8. \( \frac { 1 }{\sqrt {9x^2-7}} \)
Answer:
To integrate \( \frac { 1 }{\sqrt {9x^2-7}} \) with respect to x, we first adjust the expression inside the square root to fit a standard integral formula. We factor out the coefficient of \(x^2\).
\[ \int \frac { 1 }{\sqrt {9x^2-7}} dx \] Factor out 9 from inside the square root:
\[ = \int \frac { 1 }{\sqrt {9 \left( x^2 - \frac{7}{9} \right)}} dx \] \[ = \int \frac { 1 }{3 \sqrt {x^2 - \left( \frac{\sqrt{7}}{3} \right)^2}} dx \] \[ = \frac { 1 }{3} \int \frac { 1 }{\sqrt {x^2 - \left( \frac{\sqrt{7}}{3} \right)^2}} dx \] We use the standard integral formula: \( \int \frac { 1 }{\sqrt{x^2 - a^2}} dx = \log \left| x + \sqrt{x^2 - a^2} \right| + c \).
Here, \( a = \frac{\sqrt{7}}{3} \).
\[ = \frac { 1 }{3} \log \left| x + \sqrt{x^2 - \left( \frac{\sqrt{7}}{3} \right)^2} \right| + c \] \[ = \frac { 1 }{3} \log \left| x + \sqrt{x^2 - \frac{7}{9}} \right| + c \] This is a good example of how to handle square roots in the denominator.
To make the answer match the original denominator form more closely, we can rewrite \( \sqrt{x^2 - \frac{7}{9}} \):
\( \sqrt{x^2 - \frac{7}{9}} = \sqrt{\frac{9x^2-7}{9}} = \frac{\sqrt{9x^2-7}}{3} \)
So, the expression inside the logarithm becomes:
\( x + \frac{\sqrt{9x^2-7}}{3} = \frac{3x + \sqrt{9x^2-7}}{3} \)
Therefore, the integral is:
\[ = \frac { 1 }{3} \log \left| \frac{3x + \sqrt{9x^2-7}}{3} \right| + c \] Using logarithm properties, \( \log \left| \frac{A}{B} \right| = \log |A| - \log |B| \), so \( \frac{1}{3} \log \left| \frac{3x + \sqrt{9x^2-7}}{3} \right| + c = \frac{1}{3} \log \left| 3x + \sqrt{9x^2-7} \right| - \frac{1}{3} \log 3 + c \).
Since \( -\frac{1}{3} \log 3 \) is a constant, we can absorb it into the integration constant C.
So, the simplified form is:
\[ = \frac { 1 }{3} \log \left| 3x + \sqrt{9x^2-7} \right| + c \]In simple words: First, we take out the number 9 from inside the square root to make the expression look like a standard integral formula. Then, we use that formula, which gives us a logarithm as the answer. We then simplify the constant term.

🎯 Exam Tip: When dealing with integrals involving \( \sqrt{ax^2 \pm b} \), always factor out 'a' first to simplify the expression to \( \sqrt{a}\sqrt{x^2 \pm c} \) before applying standard formulas.

Question 9. \( \frac { 1 }{\sqrt {x^2+16x+13}} \)
Answer:
To integrate \( \frac { 1 }{\sqrt {x^2+16x+13}} \) with respect to x, we complete the square in the denominator inside the square root. This helps us use a standard integral formula.
Consider the expression inside the square root: \( x^2+16x+13 \).
To complete the square for \( x^2+16x \), we add and subtract \( \left( \frac{16}{2} \right)^2 = 8^2 = 64 \):
\( x^2+16x+13 = x^2+16x+64-64+13 \)
\( = (x+8)^2 - 51 \)
This can be written as \( (x+8)^2 - (\sqrt{51})^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{\sqrt {(x+8)^2 - (\sqrt{51})^2}} dx \] We use the standard integral formula: \( \int \frac { 1 }{\sqrt{x^2 - a^2}} dx = \log \left| x + \sqrt{x^2 - a^2} \right| + c \).
Here, \( x \) is replaced by \( (x+8) \) and \( a = \sqrt{51} \).
\[ = \log \left| (x+8) + \sqrt{(x+8)^2 - (\sqrt{51})^2} \right| + c \] Substitute back the original expression for the term inside the square root:
\[ = \log \left| (x+8) + \sqrt{x^2+16x+13} \right| + c \] Completing the square is crucial for transforming these types of expressions.
In simple words: We change the expression inside the square root by making it a perfect square minus a number. This allows us to use a special integration rule directly, giving us a logarithm involving the original and modified expressions.

🎯 Exam Tip: When completing the square inside a square root, ensure you correctly handle the constant term. If it results in a negative number, the form will be \( \sqrt{x^2-a^2} \). If it's positive, it will be \( \sqrt{x^2+a^2} \).

Question 10. \( \frac { 1 }{ \sqrt{x^2-3x+2} } \)
Answer:
To integrate \( \frac { 1 }{ \sqrt{x^2-3x+2} } \) with respect to x, we first complete the square in the denominator inside the square root. This prepares the expression for a standard integration formula.
Consider the expression inside the square root: \( x^2-3x+2 \).
To complete the square for \( x^2-3x \), we add and subtract \( \left( \frac{-3}{2} \right)^2 = \frac{9}{4} \):
\( x^2-3x+2 = x^2-3x+\frac{9}{4}-\frac{9}{4}+2 \)
\( = \left( x-\frac{3}{2} \right)^2 - \frac{9}{4} + \frac{8}{4} \)
\( = \left( x-\frac{3}{2} \right)^2 - \frac{1}{4} \)
This can be written as \( \left( x-\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \).
So, the integral becomes:
\[ \int \frac { 1 }{ \sqrt{\left( x-\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2} } dx \] We use the standard integral formula: \( \int \frac { 1 }{\sqrt{x^2 - a^2}} dx = \log \left| x + \sqrt{x^2 - a^2} \right| + c \).
Here, \( x \) is replaced by \( \left( x-\frac{3}{2} \right) \) and \( a = \frac{1}{2} \).
\[ = \log \left| \left( x-\frac{3}{2} \right) + \sqrt{\left( x-\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2} \right| + c \] Substitute back the original expression for the term inside the square root:
\[ = \log \left| \left( x-\frac{3}{2} \right) + \sqrt{x^2-3x+2} \right| + c \] This method simplifies complex integrals into recognizable forms.
In simple words: We change the math inside the square root at the bottom of the fraction to a perfect square minus a small number. Then, we use a known rule for integrals, which gives us a logarithm as the final answer.

🎯 Exam Tip: When completing the square for expressions like \(x^2+bx+c\), the constant to add and subtract is always \( (b/2)^2 \). Remember to simplify the constant term carefully.

Question 11. \( \frac {x^3 }{ \sqrt{x^8-1} } \)
Answer:
To integrate \( \frac {x^3 }{ \sqrt{x^8-1} } \) with respect to x, we use a substitution method to simplify the expression.
Let \( z = x^4 \).
Then, differentiate \( z \) with respect to \( x \): \( \frac{dz}{dx} = 4x^3 \).
This means \( dz = 4x^3 dx \), or \( \frac{1}{4} dz = x^3 dx \).
Also, \( x^8 = (x^4)^2 = z^2 \).
Substitute these into the integral:
\[ \int \frac {x^3 }{ \sqrt{x^8-1} } dx = \int \frac { \frac{1}{4} dz }{ \sqrt{z^2-1} } \] \[ = \frac{1}{4} \int \frac { 1 }{ \sqrt{z^2-1^2} } dz \] We use the standard integral formula: \( \int \frac { 1 }{\sqrt{x^2 - a^2}} dx = \log \left| x + \sqrt{x^2 - a^2} \right| + c \).
Here, \( z \) is the variable and \( a=1 \).
\[ = \frac{1}{4} \log \left| z + \sqrt{z^2-1^2} \right| + c \] Now, substitute back \( z = x^4 \):
\[ = \frac{1}{4} \log \left| x^4 + \sqrt{(x^4)^2-1} \right| + c \] \[ = \frac{1}{4} \log \left| x^4 + \sqrt{x^8-1} \right| + c \] The substitution method effectively simplifies complex-looking integrals.
In simple words: We use a trick where we let \(z\) be equal to \(x^4\). This changes the integral into a simpler form that we can solve using a basic logarithm rule. Finally, we put \(x^4\) back in place of \(z\) to get the answer.

🎯 Exam Tip: For integrals where the numerator is a derivative of a part of the denominator (like \(x^3\) for \(x^4\) or \(x^8\)), a u-substitution is usually the most efficient approach.

Question 12. \( \sqrt { 1 + x + x^2} \)
Answer:
To integrate \( \sqrt { 1 + x + x^2} \) with respect to x, we complete the square inside the square root to transform it into a standard integral form.
Consider the expression inside the square root: \( x^2+x+1 \).
To complete the square for \( x^2+x \), we add and subtract \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \):
\( x^2+x+1 = x^2+x+\frac{1}{4}-\frac{1}{4}+1 \)
\( = \left( x+\frac{1}{2} \right)^2 - \frac{1}{4} + \frac{4}{4} \)
\( = \left( x+\frac{1}{2} \right)^2 + \frac{3}{4} \)
This can be written as \( \left( x+\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 \).
So, the integral becomes:
\[ \int \sqrt{\left( x+\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} dx \] We use the standard integral formula: \( \int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + c \).
Here, \( x \) is replaced by \( \left( x+\frac{1}{2} \right) \) and \( a = \frac{\sqrt{3}}{2} \).
Let \( X = x+\frac{1}{2} \) and \( A = \frac{\sqrt{3}}{2} \).
\[ = \frac{X}{2}\sqrt{X^2+A^2} + \frac{A^2}{2}\log|X+\sqrt{X^2+A^2}| + c \] Substitute back \( X \) and \( A \):
\[ = \frac{\left( x+\frac{1}{2} \right)}{2}\sqrt{\left( x+\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{\left( \frac{\sqrt{3}}{2} \right)^2}{2}\log\left|\left( x+\frac{1}{2} \right)+\sqrt{\left( x+\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\right| + c \] Simplify the terms:
\( \left( x+\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 = x^2+x+1 \)
\( \frac{\left( \frac{\sqrt{3}}{2} \right)^2}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{8} \)
So, the integral is:
\[ = \frac{\left( x+\frac{1}{2} \right)}{2}\sqrt{x^2+x+1} + \frac{3}{8}\log\left|\left( x+\frac{1}{2} \right)+\sqrt{x^2+x+1}\right| + c \] This is a comprehensive solution for integrals of square roots involving quadratic expressions.
In simple words: We first change the expression under the square root into a perfect square plus a number. Then, we use a special integration rule for square roots that gives us a two-part answer: one part with the square root and another with a logarithm.

🎯 Exam Tip: When using the integral formula for \( \sqrt{x^2 \pm a^2} \), remember to correctly substitute the entire quadratic expression \( (x+k) \) for \(x\) and its corresponding \(a^2\) value.

Question 13. \( \sqrt {x^2 -2} \)
Answer:
To integrate \( \sqrt {x^2 -2} \) with respect to x, we recognize that it is in a standard form directly. We can rewrite 2 as \( (\sqrt{2})^2 \).
\[ \int \sqrt {x^2 - (\sqrt{2})^2} dx \] We use the standard integral formula: \( \int \sqrt{x^2 - a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + c \).
Here, \( a = \sqrt{2} \).
\[ = \frac{x}{2}\sqrt{x^2-(\sqrt{2})^2} - \frac{(\sqrt{2})^2}{2}\log|x+\sqrt{x^2-(\sqrt{2})^2}| + c \] Simplify the terms:
\[ = \frac{x}{2}\sqrt{x^2-2} - \frac{2}{2}\log|x+\sqrt{x^2-2}| + c \] \[ = \frac{x}{2}\sqrt{x^2-2} - \log|x+\sqrt{x^2-2}| + c \] This demonstrates a direct application of a common integral formula.
In simple words: We can directly use a known formula for integrating square roots that look like \(x^2\) minus a number. The formula gives us an answer that includes both a square root part and a logarithm part.

🎯 Exam Tip: Always identify the standard integral form (e.g., \( \sqrt{x^2-a^2} \), \( \sqrt{a^2-x^2} \), or \( \sqrt{x^2+a^2} \)) correctly before applying the respective formula. A mismatch will lead to an incorrect solution.

Question 14. \( \sqrt { 4x^2 -5} \)
Answer:
To integrate \( \sqrt { 4x^2 -5} \) with respect to x, we first adjust the expression inside the square root to fit a standard integral formula. We factor out the coefficient of \(x^2\).
\[ \int \sqrt { 4x^2 -5} dx \] Factor out 4 from inside the square root:
\[ = \int \sqrt { 4 \left( x^2 - \frac{5}{4} \right)} dx \] \[ = \int 2 \sqrt { x^2 - \left( \frac{\sqrt{5}}{2} \right)^2} dx \] \[ = 2 \int \sqrt { x^2 - \left( \frac{\sqrt{5}}{2} \right)^2} dx \] We use the standard integral formula: \( \int \sqrt{x^2 - a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + c \).
Here, \( a = \frac{\sqrt{5}}{2} \).
\[ = 2 \left[ \frac{x}{2}\sqrt{x^2 - \left( \frac{\sqrt{5}}{2} \right)^2} - \frac{\left( \frac{\sqrt{5}}{2} \right)^2}{2}\log\left|x+\sqrt{x^2 - \left( \frac{\sqrt{5}}{2} \right)^2}\right| \right] + c \] Simplify the terms:
\[ = 2 \left[ \frac{x}{2}\sqrt{x^2 - \frac{5}{4}} - \frac{\frac{5}{4}}{2}\log\left|x+\sqrt{x^2 - \frac{5}{4}}\right| \right] + c \] \[ = 2 \left[ \frac{x}{2}\sqrt{x^2 - \frac{5}{4}} - \frac{5}{8}\log\left|x+\sqrt{x^2 - \frac{5}{4}}\right| \right] + c \] Now, distribute the 2:
\[ = x\sqrt{x^2 - \frac{5}{4}} - \frac{5}{4}\log\left|x+\sqrt{x^2 - \frac{5}{4}}\right| + c \] We can simplify \( \sqrt{x^2 - \frac{5}{4}} = \sqrt{\frac{4x^2-5}{4}} = \frac{\sqrt{4x^2-5}}{2} \).
Substitute this back:
\[ = x \frac{\sqrt{4x^2-5}}{2} - \frac{5}{4}\log\left|x+\frac{\sqrt{4x^2-5}}{2}\right| + c \] \[ = \frac{x\sqrt{4x^2-5}}{2} - \frac{5}{4}\log\left|\frac{2x+\sqrt{4x^2-5}}{2}\right| + c \] Using logarithm properties, \( \log \left| \frac{A}{B} \right| = \log |A| - \log |B| \), we can absorb \( -\frac{5}{4}\log 2 \) into the constant C.
\[ = \frac{x\sqrt{4x^2-5}}{2} - \frac{5}{4}\log\left|2x+\sqrt{4x^2-5}\right| + c \] This solution demonstrates proper handling of coefficients and algebraic simplification.
In simple words: First, we take out the number 4 from inside the square root. This makes the expression fit a standard integral rule. We then use that rule, simplify the terms, and combine the numbers to get the final answer.

🎯 Exam Tip: Remember to simplify the square root term \( \sqrt{x^2 - \frac{a^2}{b}} \) into \( \frac{\sqrt{bx^2-a^2}}{\sqrt{b}} \) to match the original form in the final answer if possible, and adjust the constant term of the logarithm accordingly.

Question 15. \( \sqrt {2x^2 +4x+1} \)
Answer:
To integrate \( \sqrt {2x^2 +4x+1} \) with respect to x, we first simplify the expression by factoring out the coefficient of \(x^2\) and then completing the square.
Consider the expression inside the square root: \( 2x^2 +4x+1 \).
Factor out 2 from the expression:
\( 2x^2 +4x+1 = 2(x^2+2x+\frac{1}{2}) \)
So, the integral becomes:
\[ \int \sqrt {2(x^2+2x+\frac{1}{2})} dx = \sqrt{2} \int \sqrt {x^2+2x+\frac{1}{2}} dx \] Now, complete the square for \( x^2+2x+\frac{1}{2} \).
To complete the square for \( x^2+2x \), we add and subtract \( \left( \frac{2}{2} \right)^2 = 1^2 = 1 \):
\( x^2+2x+\frac{1}{2} = x^2+2x+1-1+\frac{1}{2} \)
\( = (x+1)^2 - \frac{1}{2} \)
This can be written as \( (x+1)^2 - \left( \frac{1}{\sqrt{2}} \right)^2 \).
So, the integral becomes:
\[ \sqrt{2} \int \sqrt {(x+1)^2 - \left( \frac{1}{\sqrt{2}} \right)^2} dx \] We use the standard integral formula: \( \int \sqrt{x^2 - a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + c \).
Here, \( x \) is replaced by \( (x+1) \) and \( a = \frac{1}{\sqrt{2}} \).
Let \( X = x+1 \) and \( A = \frac{1}{\sqrt{2}} \).
\[ = \sqrt{2} \left[ \frac{X}{2}\sqrt{X^2-A^2} - \frac{A^2}{2}\log|X+\sqrt{X^2-A^2}| \right] + c \] Substitute back \( X \) and \( A \):
\[ = \sqrt{2} \left[ \frac{(x+1)}{2}\sqrt{(x+1)^2 - \left( \frac{1}{\sqrt{2}} \right)^2} - \frac{\left( \frac{1}{\sqrt{2}} \right)^2}{2}\log\left|(x+1)+\sqrt{(x+1)^2 - \left( \frac{1}{\sqrt{2}} \right)^2}\right| \right] + c \] Simplify the terms:
\( (x+1)^2 - \left( \frac{1}{\sqrt{2}} \right)^2 = x^2+2x+\frac{1}{2} \)
\( \frac{\left( \frac{1}{\sqrt{2}} \right)^2}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \)
So, the expression inside the brackets is:
\[ \frac{(x+1)}{2}\sqrt{x^2+2x+\frac{1}{2}} - \frac{1}{4}\log\left|(x+1)+\sqrt{x^2+2x+\frac{1}{2}}\right| \] Now, distribute the \( \sqrt{2} \):
\[ = \frac{\sqrt{2}(x+1)}{2}\sqrt{x^2+2x+\frac{1}{2}} - \frac{\sqrt{2}}{4}\log\left|(x+1)+\sqrt{x^2+2x+\frac{1}{2}}\right| + c \] We can also rewrite \( \sqrt{2}\sqrt{x^2+2x+\frac{1}{2}} = \sqrt{2(x^2+2x+\frac{1}{2})} = \sqrt{2x^2+4x+1} \).
So, the first term becomes:
\[ \frac{(x+1)\sqrt{2x^2+4x+1}}{2} \] And the second term:
\[ = - \frac{\sqrt{2}}{4}\log\left|(x+1)+\frac{\sqrt{2x^2+4x+1}}{\sqrt{2}}\right| + c \] \[ = \frac{(x+1)\sqrt{2x^2+4x+1}}{2} - \frac{\sqrt{2}}{4}\log\left|\frac{\sqrt{2}(x+1)+\sqrt{2x^2+4x+1}}{\sqrt{2}}\right| + c \] Absorbing the \( -\frac{\sqrt{2}}{4}\log\sqrt{2} \) into the constant C:
\[ = \frac{(x+1)\sqrt{2x^2+4x+1}}{2} - \frac{\sqrt{2}}{4}\log\left|\sqrt{2}(x+1)+\sqrt{2x^2+4x+1}\right| + c \] This type of integral requires careful algebraic manipulation at each step.
In simple words: First, we take out the number 2 from under the square root. Then, we make the remaining expression a perfect square minus a number. After that, we use a special integration rule for square roots, simplify the terms, and combine everything to get the answer.

🎯 Exam Tip: When simplifying the final answer, try to express terms under the square root in their original form \( (2x^2+4x+1) \) to make the solution clearer and more concise.