int frac{(x - 3)e^x}{(x - 1)^3} , dx is equal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x - 3}{(x - 1)^3} \, dx$. Rewrite the numerator

Vedclass · Accepted Answer

$\frac{e^x}{(x - 1)^2} + c$

Vedclass · Answer

(B) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x - 3}{(x - 1)^3} \, dx$. Rewrite the numerator as $(x - 1) - 2$: $I = \int e^x \frac{(x - 1) - 2}{(x - 1)^3} \, dx$ $I = \int e^x \left[ \frac{x - 1}{(x - 1)^3} - \frac{2}{(x - 1)^3} \right] \, dx$ $I = \int e^x \left[ \frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3} \right] \, dx$ Let $f(x) = \frac{1}{(x - 1)^2} = (x - 1)^{-2}$. Then $f'(x) = -2(x - 1)^{-3} = -\frac{2}{(x - 1)^3}$. Since the integral is in the form $\int e^x [f(x) + f'(x)] \, dx$,the result is $e^x f(x) + c$. Therefore,$I = \frac{e^x}{(x - 1)^2} + c$.

$\int \frac{(x - 3)e^x}{(x - 1)^3} \, dx$ is equal to

Similar Questions

The value of the integral $\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x$ is

$\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$,where $C$ is a constant,then $\frac{d^{3} f}{d x^{3}}$ at $x = 1$ is equal to

$\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = $ . . . . . . $+ c$.

Integrate the function: $\frac{2+\sin 2x}{1+\cos 2x} e^x$

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module