int e^x frac{x^2+1}{(x+1)^2} d x is equal to | vedclass

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

Question

(C) Let $I = \int e^x \frac{x^2+1}{(x+1)^2} d x$. We can rewrite the numerator as $x^2 - 1 + 2$. $I = \int e^x \left( \frac{x^2-1+2}{(x+1)^2} \right)

Vedclass · Accepted Answer

$e^x\left(\frac{x-1}{x+1}\right)+C$

Vedclass · Answer

(C) Let $I = \int e^x \frac{x^2+1}{(x+1)^2} d x$. We can rewrite the numerator as $x^2 - 1 + 2$. $I = \int e^x \left( \frac{x^2-1+2}{(x+1)^2} \right) d x = \int e^x \left( \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2} \right) d x$. $I = \int e^x \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right) d x$. Let $f(x) = \frac{x-1}{x+1}$. Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$. Using the standard integral formula $\int e^x \{f(x) + f'(x)\} d x = e^x f(x) + C$,we get: $I = e^x \left( \frac{x-1}{x+1} \right) + C$.

$\int e^x \frac{x^2+1}{(x+1)^2} d x$ is equal to

Similar Questions

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

If $\int e^x(\sin^2 2x - 8 \cos 4x) dx = e^x f(x) + c$, then $f(\frac{\pi}{4}) = $

Integrate the function: $\frac{(x-3) e^{x}}{(x-1)^{3}}$

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is

$\int {{e^x}(1 + \tan x + {{\tan }^2}x)\,dx = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module