Practicing Success
If $\int \frac{e^x-1}{e^x+1} d x=f(x)+C$, then $f(x)=$ |
$2 \log \left(e^x+1\right)+C$ $\log \left(e^{2 x}-1\right)+C$ $2 \log \left(e^x+1\right)-x+C$ $\log \left(e^{2 x}+1\right)+C$ |
$2 \log \left(e^x+1\right)-x+C$ |
We have, $\int \frac{e^x-1}{e^x+1} d x$ $\Rightarrow \quad=\int \frac{e^x}{e^x+1} d x-\int \frac{1}{e^x+1} d x$ $ \Rightarrow \quad=\int \frac{1}{e^x+1} d\left(e^x+1\right)+\int \frac{1}{e^{-x}+1} d\left(e^{-x}+1\right)$ $ $\Rightarrow \quad=\log \left(e^x+1\right)+\log \left(e^{-x}+1\right)+C$ $\Rightarrow \quad=\log \left(e^x+1\right)+\log \left(\frac{e^x+1}{e^x}\right)+C$ $\Rightarrow \quad=\log \left(e^x+1\right)^2-\log e^x+C=2 \log \left(e^x+1\right)-x+C$ |