$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \Rightarrow \quad\left(e x^{e-1}+e^x\right) d x=d t \\ & \Rightarrow \quad e\left(x^{e-1}+\frac{e^x}{e}\right) d x=d t \\ & \Rightarrow \quad e\left(x^{e-1}+e^{x-1}\right) d x=d t\end{aligned}$

Now, putting values from Eqs. (ii) and (iii) in Eq. ( $i$ ) $ \begin{aligned} & \therefore \quad I=\frac{1}{e} \int \frac{d t}{t}=\frac{1}{e} \log |t|+C \\ & =\frac{1}{e} \log \left|x^e+e^x\right|+C \end{aligned} $

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

Solution