Browse Questions

# Choose the correct answers in $\int\frac{dx}{e^x+e^{-x}}$ is equal to

$\begin{array}{1 1} (A)\;\tan^{-1}(e^x)+C \\ (B)\;\tan^{-1}(e^{-x})+C\\C\;log(e^x-e^{-x})+C \\ (D)\;log(e^x+e^{-x})+C \end{array}$

Toolbox:
• (i)In a function $\int f(x)dx$ if f(x)=t, then $f'(x)dx=dt$ Therefore $\int f(x)dx=\int t.dt$
• (ii) $\int \frac{dx}{x^2+a^3}=\frac{1}{a} \tan ^{-1}(x/a)+c$
Given $I=\frac{dx}{e^x+e^{-x}}$

multiply and divide by $e^x$

$I=\int \frac{e^x}{e^x(e^x+e^{-x})}dx$

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

Let $e^2x=t$ on differentiating w.r.t x,

$e^xdx=dt$

Therefore $I=\int \frac{dt}{1+t^2}$

This is of the form $\int \frac{dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1}(x/a)+c$

Therefore on integrating we get

$I=\tan ^{-1}(t)+c$

substituting for t we get,

$I=\tan ^{-1}(e^x)+c$

Hence the correct answer is A