Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.11

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.11

Integrate the following with respect to x:

Question 1.
(i) \( \frac{2 x-3}{x^{2}+4 x-12} \)
Answer:
We need to integrate \( \frac{2x - 3}{x^2 + 4x - 12} \) with respect to \( x \).
First, we express the numerator in terms of the derivative of the denominator.
Let \( 2x - 3 = A \frac{\mathrm{d}}{\mathrm{d} x} (x² + 4x - 12) + B \)
\( 2x - 3 = A (2x + 4) + B \)
Comparing coefficients of \( x \): \( 2A = 2 \implies A = 1 \)
Comparing constant terms: \( 4A + B = -3 \)
Substitute \( A = 1 \): \( 4(1) + B = -3 \)
\( 4 + B = -3 \)
\( B = -3 - 4 \)
\( B = -7 \)
So, \( 2x - 3 = 1 (2x + 4) - 7 \)
Now, we split the integral into two parts:
\[ \int \frac{2x - 3}{x² + 4x - 12} dx = \int \frac{(2x + 4) - 7}{x² + 4x - 12} dx \]
\[ = \int \frac{2x + 4}{x² + 4x - 12} dx - 7 \int \frac{dx}{x² + 4x - 12} \]
For the first integral, let \( t = x² + 4x - 12 \). Then \( dt = (2x + 4)dx \).
\[ \int \frac{2x + 4}{x² + 4x - 12} dx = \int \frac{dt}{t} = \log |t| + C_1 = \log |x² + 4x - 12| + C_1 \]
For the second integral, we complete the square in the denominator:
\( x² + 4x - 12 = x² + 4x + 4 - 4 - 12 = (x + 2)² - 16 = (x + 2)² - 4² \)
Let \( u = x + 2 \). Then \( du = dx \).
\[ -7 \int \frac{dx}{x² + 4x - 12} = -7 \int \frac{du}{u² - 4²} \]
Using the formula \( \int \frac{dx}{x² - a²} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \):
\[ -7 \times \frac{1}{2 \times 4} \log \left| \frac{u - 4}{u + 4} \right| + C_2 \]
\[ = -\frac{7}{8} \log \left| \frac{(x + 2) - 4}{(x + 2) + 4} \right| + C_2 \]
\[ = -\frac{7}{8} \log \left| \frac{x - 2}{x + 6} \right| + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{2x - 3}{x² + 4x - 12} dx = \log |x² + 4x - 12| - \frac{7}{8} \log \left| \frac{x - 2}{x + 6} \right| + C \]
This method helps simplify complex fractions into easier-to-integrate forms by separating them.
In simple words: To solve this, we first rewrite the top part of the fraction using the derivative of the bottom part. This lets us split the integral into two simpler parts. One part can be solved by simple substitution, and the other by completing the square in the denominator and using a standard integration formula.

🎯 Exam Tip: When the numerator is a linear function and the denominator is a quadratic, always try to express the numerator as \( A \times (\text{derivative of denominator}) + B \). This is a common and effective technique.

Question 1.
(ii) \( \frac{5 x-2}{2+2x+x^{2}} \)
Answer:
We need to integrate \( \frac{5x - 2}{x² + 2x + 2} \) with respect to \( x \).
Let the numerator \( 5x - 2 = A \frac{\mathrm{d}}{\mathrm{d} x} (x² + 2x + 2) + B \)
\( 5x - 2 = A (2x + 2) + B \)
Comparing coefficients of \( x \): \( 2A = 5 \implies A = \frac{5}{2} \)
Comparing constant terms: \( 2A + B = -2 \)
Substitute \( A = \frac{5}{2} \): \( 2 \left( \frac{5}{2} \right) + B = -2 \)
\( 5 + B = -2 \)
\( B = -2 - 5 \)
\( B = -7 \)
So, \( 5x - 2 = \frac{5}{2} (2x + 2) - 7 \)
Now, we split the integral:
\[ \int \frac{5x - 2}{x² + 2x + 2} dx = \int \frac{\frac{5}{2}(2x + 2) - 7}{x² + 2x + 2} dx \]
\[ = \frac{5}{2} \int \frac{2x + 2}{x² + 2x + 2} dx - 7 \int \frac{dx}{x² + 2x + 2} \]
For the first integral, let \( t = x² + 2x + 2 \). Then \( dt = (2x + 2)dx \).
\[ \frac{5}{2} \int \frac{dt}{t} = \frac{5}{2} \log |t| + C_1 = \frac{5}{2} \log |x² + 2x + 2| + C_1 \]
For the second integral, complete the square in the denominator:
\( x² + 2x + 2 = x² + 2x + 1 + 1 = (x + 1)² + 1² \)
Let \( u = x + 1 \). Then \( du = dx \).
\[ -7 \int \frac{dx}{(x + 1)² + 1²} = -7 \int \frac{du}{u² + 1²} \]
Using the formula \( \int \frac{dx}{x² + a²} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \):
\[ -7 \times \frac{1}{1} \tan^{-1} \left( \frac{u}{1} \right) + C_2 = -7 \tan^{-1} (x + 1) + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{5x - 2}{x² + 2x + 2} dx = \frac{5}{2} \log |x² + 2x + 2| - 7 \tan^{-1} (x + 1) + C \]
This approach helps to systematically integrate fractions involving quadratic terms in the denominator.
In simple words: We break the integral into two parts. The first part is solved by substituting the denominator, giving a logarithm. The second part involves completing the square in the denominator to get a form that integrates to an inverse tangent function.

🎯 Exam Tip: Remember that for integrals with a quadratic denominator in the form \( ax^2 + bx + c \), completing the square is essential to transform it into the \( x^2 \pm a^2 \) form, leading to \( \log \) or \( \tan^{-1} \) functions.

Question 1.
(iii) \( \frac{3 x+1}{2x^{2}-2x+3} \)
Answer:
We need to integrate \( \frac{3x + 1}{2x² - 2x + 3} \) with respect to \( x \).
Let the numerator \( 3x + 1 = A \frac{\mathrm{d}}{\mathrm{d} x} (2x² - 2x + 3) + B \)
\( 3x + 1 = A (4x - 2) + B \)
Comparing coefficients of \( x \): \( 4A = 3 \implies A = \frac{3}{4} \)
Comparing constant terms: \( -2A + B = 1 \)
Substitute \( A = \frac{3}{4} \): \( -2 \left( \frac{3}{4} \right) + B = 1 \)
\( -\frac{3}{2} + B = 1 \)
\( B = 1 + \frac{3}{2} \)
\( B = \frac{5}{2} \)
So, \( 3x + 1 = \frac{3}{4} (4x - 2) + \frac{5}{2} \)
Now, we split the integral:
\[ \int \frac{3x + 1}{2x² - 2x + 3} dx = \int \frac{\frac{3}{4}(4x - 2) + \frac{5}{2}}{2x² - 2x + 3} dx \]
\[ = \frac{3}{4} \int \frac{4x - 2}{2x² - 2x + 3} dx + \frac{5}{2} \int \frac{dx}{2x² - 2x + 3} \]
For the first integral, let \( t = 2x² - 2x + 3 \). Then \( dt = (4x - 2)dx \).
\[ \frac{3}{4} \int \frac{dt}{t} = \frac{3}{4} \log |t| + C_1 = \frac{3}{4} \log |2x² - 2x + 3| + C_1 \]
For the second integral, complete the square in the denominator:
\( 2x² - 2x + 3 = 2(x² - x + \frac{3}{2}) \)
\( = 2 \left( x² - x + \frac{1}{4} - \frac{1}{4} + \frac{3}{2} \right) \)
\( = 2 \left( \left( x - \frac{1}{2} \right)² + \frac{5}{4} \right) = 2 \left( \left( x - \frac{1}{2} \right)² + \left( \frac{\sqrt{5}}{2} \right)² \right) \)
Now, the second integral becomes:
\[ \frac{5}{2} \int \frac{dx}{2 \left( \left( x - \frac{1}{2} \right)² + \left( \frac{\sqrt{5}}{2} \right)² \right)} = \frac{5}{4} \int \frac{dx}{\left( x - \frac{1}{2} \right)² + \left( \frac{\sqrt{5}}{2} \right)²} \]
Let \( u = x - \frac{1}{2} \). Then \( du = dx \).
\[ \frac{5}{4} \int \frac{du}{u² + \left( \frac{\sqrt{5}}{2} \right)²} \]
Using the formula \( \int \frac{dx}{x² + a²} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \):
\[ \frac{5}{4} \times \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1} \left( \frac{u}{\frac{\sqrt{5}}{2}} \right) + C_2 \]
\[ = \frac{5}{4} \times \frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}} \right) + C_2 \]
\[ = \frac{5}{2\sqrt{5}} \tan^{-1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C_2 \]
\[ = \frac{\sqrt{5}}{2} \tan^{-1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{3x + 1}{2x² - 2x + 3} dx = \frac{3}{4} \log |2x² - 2x + 3| + \frac{\sqrt{5}}{2} \tan^{-1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C \]
This shows how to handle integrals with a coefficient for the \( x^2 \) term by factoring it out.
In simple words: First, we write the top part of the fraction using the derivative of the bottom part. Then we split the integral. For the second part, we need to factor out the number in front of \( x^2 \) and then complete the square to use the inverse tangent formula.

🎯 Exam Tip: When the denominator has a coefficient other than 1 for \( x^2 \), remember to factor out that coefficient before completing the square to apply standard integral formulas correctly.

Question 2.
(i) \( \frac{2 x+1}{\sqrt{9+4 x-x^{2}}} \)
Answer:
We need to integrate \( \frac{2x + 1}{\sqrt{9 + 4x - x²}} \) with respect to \( x \).
Let the numerator \( 2x + 1 = A \frac{\mathrm{d}}{\mathrm{d} x} (9 + 4x - x²) + B \)
\( 2x + 1 = A (4 - 2x) + B \)
Comparing coefficients of \( x \): \( -2A = 2 \implies A = -1 \)
Comparing constant terms: \( 4A + B = 1 \)
Substitute \( A = -1 \): \( 4(-1) + B = 1 \)
\( -4 + B = 1 \)
\( B = 1 + 4 \)
\( B = 5 \)
So, \( 2x + 1 = -1 (4 - 2x) + 5 \)
Now, we split the integral:
\[ \int \frac{2x + 1}{\sqrt{9 + 4x - x²}} dx = \int \frac{-(4 - 2x) + 5}{\sqrt{9 + 4x - x²}} dx \]
\[ = - \int \frac{4 - 2x}{\sqrt{9 + 4x - x²}} dx + 5 \int \frac{dx}{\sqrt{9 + 4x - x²}} \]
For the first integral, let \( t = 9 + 4x - x² \). Then \( dt = (4 - 2x)dx \).
\[ - \int \frac{dt}{\sqrt{t}} = - \int t^{-1/2} dt = - \frac{t^{1/2}}{1/2} + C_1 = -2\sqrt{t} + C_1 = -2\sqrt{9 + 4x - x²} + C_1 \]
For the second integral, complete the square in the denominator under the square root:
\( 9 + 4x - x² = 9 - (x² - 4x) \)
\( = 9 - (x² - 4x + 4 - 4) \)
\( = 9 - ((x - 2)² - 4) \)
\( = 9 - (x - 2)² + 4 \)
\( = 13 - (x - 2)² = (\sqrt{13})² - (x - 2)² \)
Let \( u = x - 2 \). Then \( du = dx \).
\[ 5 \int \frac{dx}{\sqrt{(\sqrt{13})² - (x - 2)²}} = 5 \int \frac{du}{\sqrt{(\sqrt{13})² - u²}} \]
Using the formula \( \int \frac{dx}{\sqrt{a² - x²}} = \sin^{-1} \left( \frac{x}{a} \right) + C \):
\[ 5 \sin^{-1} \left( \frac{u}{\sqrt{13}} \right) + C_2 = 5 \sin^{-1} \left( \frac{x - 2}{\sqrt{13}} \right) + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{2x + 1}{\sqrt{9 + 4x - x²}} dx = -2\sqrt{9 + 4x - x²} + 5 \sin^{-1} \left( \frac{x - 2}{\sqrt{13}} \right) + C \]
This method is crucial for integrals with a linear term in the numerator and a quadratic term under a square root in the denominator.
In simple words: We rewrite the top part of the fraction using the derivative of the expression under the square root. This splits the problem into two easier integrals. One is a direct substitution leading to a square root, and the other requires completing the square to use the inverse sine formula.

🎯 Exam Tip: When dealing with \( \sqrt{a^2 - x^2} \), remember the \( \sin^{-1} \) integral. Always be careful when factoring out a negative sign during completing the square for such expressions.

Question 2.
(ii) \( \frac{x+2}{\sqrt{x^{2}-1}} \)
Answer:
We need to integrate \( \frac{x + 2}{\sqrt{x² - 1}} \) with respect to \( x \).
Let the numerator \( x + 2 = A \frac{\mathrm{d}}{\mathrm{d} x} (x² - 1) + B \)
\( x + 2 = A (2x) + B \)
Comparing coefficients of \( x \): \( 2A = 1 \implies A = \frac{1}{2} \)
Comparing constant terms: \( B = 2 \)
So, \( x + 2 = \frac{1}{2} (2x) + 2 \)
Now, we split the integral:
\[ \int \frac{x + 2}{\sqrt{x² - 1}} dx = \int \frac{\frac{1}{2}(2x) + 2}{\sqrt{x² - 1}} dx \]
\[ = \frac{1}{2} \int \frac{2x}{\sqrt{x² - 1}} dx + 2 \int \frac{dx}{\sqrt{x² - 1}} \]
For the first integral, let \( t = x² - 1 \). Then \( dt = 2xdx \).
\[ \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \frac{t^{1/2}}{1/2} + C_1 = \sqrt{t} + C_1 = \sqrt{x² - 1} + C_1 \]
For the second integral, this is a standard formula:
\[ 2 \int \frac{dx}{\sqrt{x² - 1²}} = 2 \log |x + \sqrt{x² - 1²}| + C_2 = 2 \log |x + \sqrt{x² - 1}| + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{x + 2}{\sqrt{x² - 1}} dx = \sqrt{x² - 1} + 2 \log |x + \sqrt{x² - 1}| + C \]
This problem demonstrates how to separate an integral into two parts that can be solved using substitution and a direct formula, respectively.
In simple words: We break the fraction into two parts. The first part is easy to solve by changing the variable, which gives a square root. The second part is a direct formula for integrating fractions with square roots like \( \sqrt{x^2 - a^2} \), resulting in a logarithm.

🎯 Exam Tip: Always look for terms that are derivatives of the expression under the square root. The integral of \( \frac{f'(x)}{\sqrt{f(x)}} \) is \( 2\sqrt{f(x)} \).

Question 2.
(iii) \( \frac{2 x+3}{\sqrt{x^{2}+4x+1}} \)
Answer:
We need to integrate \( \frac{2x + 3}{\sqrt{x² + 4x + 1}} \) with respect to \( x \).
Let the numerator \( 2x + 3 = A \frac{\mathrm{d}}{\mathrm{d} x} (x² + 4x + 1) + B \)
\( 2x + 3 = A (2x + 4) + B \)
Comparing coefficients of \( x \): \( 2A = 2 \implies A = 1 \)
Comparing constant terms: \( 4A + B = 3 \)
Substitute \( A = 1 \): \( 4(1) + B = 3 \)
\( 4 + B = 3 \)
\( B = 3 - 4 \)
\( B = -1 \)
So, \( 2x + 3 = 1 (2x + 4) - 1 \)
Now, we split the integral:
\[ \int \frac{2x + 3}{\sqrt{x² + 4x + 1}} dx = \int \frac{(2x + 4) - 1}{\sqrt{x² + 4x + 1}} dx \]
\[ = \int \frac{2x + 4}{\sqrt{x² + 4x + 1}} dx - \int \frac{dx}{\sqrt{x² + 4x + 1}} \]
For the first integral, let \( t = x² + 4x + 1 \). Then \( dt = (2x + 4)dx \).
\[ \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt = \frac{t^{1/2}}{1/2} + C_1 = 2\sqrt{t} + C_1 = 2\sqrt{x² + 4x + 1} + C_1 \]
For the second integral, complete the square in the denominator under the square root:
\( x² + 4x + 1 = x² + 4x + 4 - 4 + 1 = (x + 2)² - 3 = (x + 2)² - (\sqrt{3})² \)
Let \( u = x + 2 \). Then \( du = dx \).
\[ - \int \frac{dx}{\sqrt{(x + 2)² - (\sqrt{3})²}} = - \int \frac{du}{\sqrt{u² - (\sqrt{3})²}} \]
Using the formula \( \int \frac{dx}{\sqrt{x² - a²}} = \log |x + \sqrt{x² - a²}| + C \):
\[ - \log |u + \sqrt{u² - (\sqrt{3})²}| + C_2 \]
\[ = - \log |(x + 2) + \sqrt{(x + 2)² - (\sqrt{3})²}| + C_2 \]
\[ = - \log |x + 2 + \sqrt{x² + 4x + 1}| + C_2 \]
Combining both results, the final integral is:
\[ \int \frac{2x + 3}{\sqrt{x² + 4x + 1}} dx = 2\sqrt{x² + 4x + 1} - \log |x + 2 + \sqrt{x² + 4x + 1}| + C \]
This example shows how to combine substitution and completing the square for integration problems involving square roots.
In simple words: First, we change the top part of the fraction to use the derivative of the expression inside the square root. This makes two separate integrals. One integral is solved by simple substitution, and the other requires completing the square inside the square root to use a logarithm formula.

🎯 Exam Tip: When completing the square for expressions like \( x^2 + 4x + 1 \), ensure the constant term \( (+4-4) \) is correctly handled to match the general form \( (x \pm k)^2 \pm a^2 \).