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

$\int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} \,d x$ is equal to

$\int e^{x \operatorname{cosec} x} \cdot \operatorname{cosec} x \cdot(1-x \cot x) \, dx =$

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

If $\int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$,then $k = $ . . . . . . .

Evaluate the definite integral $\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module