Question

$\int \frac{a^x+{\tan }^{-1}{a}^x}{1+a^{2x}}dx$

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

Answer 1

The correct option is A $\frac{a^{{\tan }^{-1}{a}^x}}{{\left(\log a\right)}^2}$

Let $I=\int \frac{a^x+{\tan }^{-1}{a}^x}{1+a^{2x}}dx$
Put $a^x=t\Rightarrow a^x\log adx=dt$
Therefore
$I=\int \frac{a^x.a^{{\tan }^{-1}{a}^x}}{1+a^{2x}}dx=\int \frac{a^{{\tan }^{-1}t}t}{(1+t^2)}\frac{dt}{\log a}$
Now put ${\tan }^{-1}t=z\Rightarrow \frac{1}{1+t^2}dt=dz$
Therefore
$I=\int \frac{a^{z}dz}{\log a}=\frac{a^z}{{\left(\log a\right)}^2}=\frac{a^{{\tan }^{-1}{a}^x}}{{\left(\log a\right)}^2}$
Hence, option 'A' is correct.