Definite Integration
Integration by parts
Grade Class 12
Question:
$\\int e^x \left( \frac{x^2 - 3}{(x - 1)^2} \right) dx$ is equal to -(where $C$ is constant of integration)
$e^x \left( \frac{x + 3}{x - 1} \right) + C$
$e^x \left( \frac{x - 3}{x - 1} \right) + C$
$e^x \left( \frac{x + 1}{x - 1} \right) + C$
$e^x \left( \frac{1}{x - 1} \right)^2 + C$
Step-by-Step Solution
Key Concept: The integral is of the form integral e^x * (f(x) + f'(x)) dx = e^x * f(x) + C. Rewrite (x^2 - 3) / (x - 1)^2 as (x^2 - 1 - 2) / (x - 1)^2 = (x + 1)/(x - 1) - 2/(x - 1)^2. Let f(x) = (x + 1)/(x - 1). Then f'(x) = ((x - 1) - (x + 1)) / (x - 1)^2 = -2 / (x - 1)^2.
We have $\int e^x \left( \frac{x^2 - 3}{(x - 1)^2} \right) dx = \int e^x \left( \frac{x^2 - 1 - 2}{(x - 1)^2} \right) dx = \int e^x \left( \frac{(x - 1)(x + 1)}{(x - 1)^2} - \frac{2}{(x - 1)^2} \right) dx = \int e^x \left( \frac{x + 1}{x - 1} - \frac{2}{(x - 1)^2} \right) dx$. Let $f(x) = \frac{x + 1}{x - 1}$. Then $f'(x) = \frac{(x - 1)(1) - (x + 1)(1)}{(x - 1)^2} = \frac{x - 1 - x - 1}{(x - 1)^2} = \frac{-2}{(x - 1)^2}$. Thus, the integral is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C = e^x \left( \frac{x + 1}{x - 1} \right) + C$.
Correct Answer: C