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

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

Question

(B) Let $I = \int \frac{e^x}{(1 + e^x)(2 + e^x)} dx$. Using partial fractions,let $\frac{e^x}{(1 + e^x)(2 + e^x)} = \frac{A}{1 + e^x} + \frac{B}{2 + e

Vedclass · Accepted Answer

$\log \left( \frac{1 + e^x}{2 + e^x} \right) + c$

Vedclass · Answer

(B) Let $I = \int \frac{e^x}{(1 + e^x)(2 + e^x)} dx$. Using partial fractions,let $\frac{e^x}{(1 + e^x)(2 + e^x)} = \frac{A}{1 + e^x} + \frac{B}{2 + e^x}$. Then $e^x = A(2 + e^x) + B(1 + e^x)$. Setting $e^x = -1$,we get $A = -1$. Setting $e^x = -2$,we get $B = 1$. So,the integral becomes $\int \left( \frac{1}{2 + e^x} - \frac{1}{1 + e^x} \right) e^x dx$. Let $u = e^x$,then $du = e^x dx$. The integral is $\int \left( \frac{1}{2 + u} - \frac{1}{1 + u} \right) du = \log|2 + u| - \log|1 + u| + c = \log \left| \frac{2 + e^x}{1 + e^x} \right| + c$. Note: The provided option $B$ is $\log \left( \frac{1 + e^x}{2 + e^x} \right) + c$,which is $-\log \left| \frac{2 + e^x}{1 + e^x} \right| + c$. Since the constant $c$ is arbitrary,the result is equivalent to $\log \left| \frac{1 + e^x}{2 + e^x} \right| + c$.

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

Similar Questions

$\int \frac{x + 2}{x^2 - 7x + 12} dx$ is equal to

$\int_{\log 4}^{\log 5} \frac{e^{2 x}+e^x}{e^{2 x}-5 e^x+6} d x=$

If $\int \frac{2x+3}{(x-1)(x^2+1)} dx = \log_e {(x-1)^{\frac{5}{2}}(x^2+1)^a} - \frac{1}{2} \tan^{-1} x + A$ where $A$ is an arbitrary constant,then the value of $a$ is

Integrate the function: $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$

$\int \frac{x+1}{x(1+x e^x)^2} \,d x=$ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module