Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.3 Solutions

Evaluate The Following:

Question 1. \( \int \frac{3e^{2t}+5}{4e^{2t}-5} dt \)
Answer:
Solution:
Let \( I = \int \frac{3e^{2t}+5}{4e^{2t}-5} dt \)
Put, Numerator \( = A(\text{Denominator}) + B\left[ \frac{d}{dt} (\text{Denominator}) \right] \)
\( \therefore 3e^{2t} + 5 = A(4e^{2t} - 5) + B\left[ \frac{d}{dt} (4e^{2t}-5) \right] \)
\( \therefore 3e^{2t} + 5 = A(4e^{2t} - 5) + B[4e^{2t} \times 2 - 0] \)
\( \therefore 3e^{2t} + 5 = (4A + 8B) e^{2t} - 5A \)
Equating the coefficient of \( e^{2t} \) and constant on both sides, we get
\( 4A + 8B = 3 \)
and \( -5A = 5 \)
\( \therefore A = -1 \)
from (1), \( 4(-1) + 8B = 3 \)
\( \therefore 8B = 7 \)
\( \therefore B = \frac{7}{8} \)
\( \therefore 3e^{2t} + 5 = -1(4e^{2t} - 5) + \frac{7}{8}(8e^{2t}) \)
\( \therefore I = \int \frac{-1(4e^{2t}-5) + \frac{7}{8}(8e^{2t})}{4e^{2t}-5} dt \)
\( = \int \left( -1 + \frac{7}{8} \frac{8e^{2t}}{4e^{2t}-5} \right) dt \)
\( = \int -1 dt + \frac{7}{8} \int \frac{8e^{2t}}{4e^{2t}-5} dt \)
\( = -t + \frac{7}{8} \log |4e^{2t}-5| + C \)
\( \left[ \therefore \int \frac{f'(x)}{f(x)} dx = \log |f(x)|+C \right] \)
In simple words: This problem involves integrating a rational function of \(e^{2t}\). We use the technique of expressing the numerator as a linear combination of the denominator and its derivative to simplify the integral into basic forms: \( \int k dt \) and \( \int \frac{f'(t)}{f(t)} dt \).

🎯 Exam Tip: Remember to equate coefficients carefully when using the method of indeterminate coefficients for integration. Pay close attention to the variable of integration (\(t\) or \(x\)) when differentiating and integrating.

Question 2. \( \int \frac{20-12e^x}{3e^x-4} dx \)
Answer:
Solution:
Let \( I = \int \frac{20-12e^x}{3e^x-4} dx \)
Put, Numerator \( = A (\text{Denominator}) + B\left[ \frac{d}{dx} (\text{Denominator}) \right] \)
\( \therefore 20 - 12e^x = A(3e^x - 4) + B\left[ \frac{d}{dx} (3e^x - 4) \right] \)
\( \therefore 20 - 12e^x = A(3e^x - 4) + B(3e^x - 0) \)
\( \therefore 20 - 12e^x = (3A + 3B)e^x - 4A \)
Equating the coefficient of \( e^x \) and constant on both sides, we get
\( 3A + 3B = -12 ......(1) \)
and \( -4A = 20 \)
\( \therefore A = -5 \)
from (1), \( 3(-5) + 3B = -12 \)
\( \therefore -15 + 3B = -12 \)
\( \therefore 3B = 3 \)
\( \therefore B = 1 \)
\( \therefore 20 - 12e^x = -5(3e^x - 4) + 1(3e^x) \)
\( \therefore I = \int \frac{-5(3e^x-4) + 3e^x}{3e^x-4} dx \)
\( = \int \left( -5 + \frac{3e^x}{3e^x-4} \right) dx \)
\( = -5 \int 1 dx + \int \frac{3e^x}{3e^x-4} dx \)
\( = -5x + \log |3e^x-4| + C \)
\( \left[ \therefore \int \frac{f'(x)}{f(x)} dx = \log |f(x)|+C \right] \)
In simple words: We decompose the rational function into two parts. The first part integrates to a linear term, and the second part matches the form of \( \int \frac{f'(x)}{f(x)} dx \), which integrates to \( \log|f(x)| \).

🎯 Exam Tip: When the numerator is a linear combination of \(e^x\) and a constant, and the denominator is also a linear combination of \(e^x\) and a constant, this substitution method simplifies the integral effectively.

Question 3. \( \int \frac{3e^x+4}{2e^x-8} dx \)
Answer:
Solution:
Let \( I = \int \frac{3e^x+4}{2e^x-8} dx \)
Put, Numerator \( = A (\text{Denominator}) + B\left[ \frac{d}{dx} (\text{Denominator}) \right] \)
\( \therefore 3e^x + 4 = A(2e^x - 8) + B\left[ \frac{d}{dx} (2e^x - 8) \right] \)
\( \therefore 3e^x + 4 = A(2e^x - 8) + B(2e^x - 0) \)
\( \therefore 3e^x + 4 = (2A + 2B)e^x - 8A \)
Equating the coefficient of \( e^x \) and constant on both sides, we get
\( 2A + 2B = 3 ........(1) \)
and \( -8A = 4 \)
\( \therefore A = -\frac{1}{2} \)
\( \therefore \) from (1), \( 2\left(-\frac{1}{2}\right) + 2B = 3 \)
\( \therefore -1 + 2B = 3 \)
\( \therefore 2B = 4 \)
\( \therefore B = 2 \)
\( \therefore 3e^x + 4 = -\frac{1}{2}(2e^x - 8) + 2(2e^x) \)
\( \therefore I = \int \frac{-\frac{1}{2}(2e^x-8)+2(2e^x)}{2e^x-8} dx \)
\( = \int \left( -\frac{1}{2} + \frac{2(2e^x)}{2e^x-8} \right) dx \)
\( = -\frac{1}{2} \int 1 dx + 2 \int \frac{2e^x}{2e^x-8} dx \)
\( = -\frac{1}{2} x + 2 \log |2e^x-8| + C \)
\( \left[ \therefore \int \frac{f'(x)}{f(x)} dx = \log |f(x)|+C \right] \)
In simple words: This problem uses a similar technique to the previous ones, where the numerator is rewritten in terms of the denominator and its derivative. This allows us to separate the integral into a simple constant term and a logarithmic term.

🎯 Exam Tip: Remember to correctly calculate the derivative of the denominator. Any error in the constants A or B will lead to an incorrect final answer.

Question 4. \( \int \frac{2e^x+5}{2e^x+1} dx \)
Answer:
Solution:
Let \( I = \int \frac{2e^x+5}{2e^x+1} dx \)
Let \( 2e^x + 5 = A(2e^x + 1) + B \frac{d}{dx} (2e^x + 1) \)
\( = A(2e^x) + A + B(2e^x) \)
\( \therefore 2e^x + 5 = (2A + 2B)e^x + A \)
Comparing the coefficients of \( e^x \) and constant term on both sides, we get
\( 2A + 2B = 2 \) and \( A = 5 \)
Solving these equations, we get
\( 2(5) + 2B = 2 \)
\( 10 + 2B = 2 \)
\( 2B = -8 \)
\( B = -4 \)
\( \therefore 2e^x + 5 = 5(2e^x + 1) - 4(2e^x) \)
\( \therefore I = \int \frac{5(2e^x + 1) - 4(2e^x)}{2e^x + 1} dx \)
\( = 5 \int 1 dx - 4 \int \frac{2e^x}{2e^x+1} dx \)
\( = 5x - 4 \log |2e^x+1| + C \)
\( \left[ \therefore \int \frac{f'(x)}{f(x)} dx = \log |f(x)|+C \right] \)
In simple words: The problem uses a substitution technique to rewrite the numerator in a form that allows us to split the integral into a simple constant integral and another integral that directly uses the \( \int \frac{f'(x)}{f(x)} dx \) formula.

🎯 Exam Tip: Pay attention to the coefficients when setting up equations for A and B. A common mistake is miscalculating the derivative of the denominator, which can affect the entire solution.