Question

Evaluate $\int \frac{e^x-2}{e^{2x}+4}dx.$

• A. $-\frac{1}{2}x+log(e^{2x}+4)-\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)$
• B. $-2x-log(e^{2x}+4)+\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)$
• C. $\frac{1}{2}x-log(e^{2x}+4)+\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)$
• D. $-x+log(e^{2x}+4)+\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)$

Answer 1

The correct option is C $\frac{1}{2}x-log(e^{2x}+4)+\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)$

Let $I=\int \frac{e^x+2}{e^{2x}+4}dx=\int (\frac{e^x}{e^{2x}+4}+\frac{2}{e^{2x}+4})dx$
$\Rightarrow I=I_1+I_2$
Where
$I_1=\int \frac{e^x}{e^{2x}+4}dx$
Put $t=e^x\Rightarrow dt=e^{x}dx$
$I_1=\int \frac{1}{t^2+4}dt=\frac{1}{2}{\tan }^{-1}\frac{t}{2}=\frac{1}{2}{\tan }^{-1}\frac{e^x}{2}$
And $I_2=\int \frac{2}{e^{2x}+4}dx$
Put $u=e^{2x}\Rightarrow du={2e}^{2x}dx$
$I_2=\int \frac{1}{u(u+4)}du=\int (\frac{1}{4u}-\frac{1}{4(u+4)})du$
$=\frac{\log u}{4}-\frac{\log (u+4)}{4}=\frac{\log e^{2x}}{4}-\frac{\log (e^{2x}+4)}{4}$
Hence
$I=\frac{\log e^{2x}}{4}-\frac{\log (e^{2x}+4)}{4}+\frac{1}{2}{\tan }^{-1}\frac{e^x}{2}$
$=\frac{1}{2}x-\frac{\log (e^{2x}+4)}{4}+\frac{1}{2}{\tan }^{-1}\frac{e^x}{2}$
Hence, option 'C' is correct.