Calculate the indefinite integral of a given ration

Question

Let $f : \mathbb{R} \rightarrow (0, \infty)$ be a differentiable function such that its derivative is continuous.

Calculate: $$\int \frac{f(x) + f'(x)}{f(x) + e^{-x}} dx$$

I need a solution which involves only the method of integration by parts.

I've tried several ways to calculate the integral, but got nothing good.

Thank you in advance!

Answer 1

Multiply all the parts by $e^x$:

$$\mathcal{I}=\int\frac{f(x)+f'(x)}{f(x)+e^{-x}}\space\text{d}x=\int\frac{f(x)e^x+f'(x)e^x}{f(x)e^x+e^{-x}e^x}\space\text{d}x=\int\frac{f(x)e^x+f'(x)e^x}{f(x)e^x+1}\space\text{d}x$$

Now, substitute $u=f(x)e^x+1$ and $\text{d}u=f(x)e^x+f'(x)e^x\space\text{d}x$:

$$\mathcal{I}=\int\frac{f(x)+f'(x)}{f(x)+e^{-x}}\space\text{d}x=\int\frac{1}{u}\space\text{d}u=\ln\left|u\right|+\text{C}=\ln\left|f(x)e^x+1\right|+\text{C}$$

Answer 2

$$\int \frac{f(x) + f'(x)}{f(x) + e^{-x}} dx=\int \frac{f(x)e^x + f'(x)e^x}{e^xf(x) +1} dx$$

Put $u=e^x f(x)+1$ then answer is $\ln \left|e^xf(x)+1\right|$