Q104 / 160 · AP EAMCET 2015 · Mathematics

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

  • A
    $\log \left|\frac{1+x e^x}{x e^x}\right|+C$
  • Correct answer
    B
    $\log \left|\frac{x e^x}{1+x e^x}\right|+C$
  • C
    $\log \left|x e^x\left(1+x e^x\right)\right|+C$
  • D
    $\log \left|1+x e^x\right|+C$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{x+1}{x\left(1+x e^x\right)} d x \\ & \text { Put } x e^x=t \\ & \Rightarrow \quad\left(x e^x+e^x\right) d x=d t \\ & \Rightarrow \quad e^x(x+1) d x=d t \\ & \Rightarrow \quad(x+1) d x=\frac{d t}{e^x} \\ & \Rightarrow \quad I=\int \frac{d t}{x e^x\left(1+x e^x\right)}=\int \frac{1}{t(1+t)} d t=\int \frac{t+1-t}{t(t+1)} d t \\ & \therefore \quad I=\int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \\ & =\log t-\log (t+1)+C \\ & =\log \left|\frac{t}{t+1}\right|+C\end{aligned}$ Now, Substitute the value of $t$, we get $=\log \left|\frac{x e^x}{x e^x+1}\right|+C$

Hence, the correct answer is option: 'B'.

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