Question

$f^{\prime }\left(x\right)=f\left(x\right),f\left(0\right)=1$, then $\int \frac{dx}{f\left(x\right)+f(-x)}$

• A. $log(e^{2x}+1)+C$
• B. $log(e^x+e^{-x})+C$
• C. $ta{n}^{-1}\left(e^x\right)+C$
• D. None

Answer 1

Answer: (C)

$ta{n}^{-1}\left(e^x\right)+C$

Given that $f^{\prime }\left(x\right)=f\left(x\right)$
$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int dx$
$\Rightarrow \ln f\left(x\right)=x+C$
when $x=0$, $f\left(0\right)=1$
Therefore, $C=0$
$f\left(x\right)=e^x$
now, $\int \frac{dx}{f\left(x\right)+f(-x)}=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{e^{x}dx}{1+e^{2x}}$

$={\tan }^{-1}{e}^x+c$
Ans: C