int frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) | vedclass

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

Question

(A) We have $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x = \int \frac{(x^{2}-1+2) e^{x}}{(x+1)^{2}} d x$.$= \int \left( \frac{(x-1)(x+1)}{(x+1)^{2}} +

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) We have $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x = \int \frac{(x^{2}-1+2) e^{x}}{(x+1)^{2}} d x$.$= \int \left( \frac{(x-1)(x+1)}{(x+1)^{2}} + \frac{2}{(x+1)^{2}} \right) e^{x} d x = \int \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^{2}} \right) e^{x} d x$.This is in the form $\int (f(x) + f'(x)) e^{x} d x = f(x) e^{x} + C$,where $f(x) = \frac{x-1}{x+1}$.We can write $f(x) = \frac{x+1-2}{x+1} = 1 - 2(x+1)^{-1}$.Then $f'(x) = 2(x+1)^{-2}$.$f''(x) = -4(x+1)^{-3}$.$f'''(x) = 12(x+1)^{-4} = \frac{12}{(x+1)^{4}}$.At $x = 1$,$f'''(1) = \frac{12}{(1+1)^{4}} = \frac{12}{16} = \frac{3}{4}$.

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

Similar Questions

$\int {{e^x}\left[ {{{\sin }^{ - 1}}\frac{x}{a} + \frac{1}{{\sqrt {{a^2} - {x^2}} }}} \right]dx = }$

$\int {{e^x}(1 + \tan x)\sec x\,dx = }$

The value of $\int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$ is

If $I = \int e^{\sin \theta} (\log \sin \theta + \operatorname{cosec}^2 \theta) \cos \theta \, d\theta$,then $I$ is equal to

Integrate the function: $e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module