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

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

Question

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d

Vedclass · Accepted Answer

$\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+c$

Vedclass · Answer

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

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

Similar Questions

$\int \frac{x^3}{x^4+3 x^2+2} d x=$

If $\int \frac{2x-1}{(x-1)(x+2)(x-3)} dx = A \log |x-1| + B \log |x+2| + C \log |x-3| + K$,then $A, B, C$ are respectively:

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

$\int \frac{x^3-1}{x^3+x} dx =$

If $\int {\frac{{(2{x^2} + 1)\,dx}}{{({x^2} - 4)({x^2} - 1)}} = \log \left[ {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^a}\,{{\left( {\frac{{x - 2}}{{x + 2}}} \right)}^b}} \right]} + C,$ then the values of $a$ and $b$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module