Question

Enter '1' if following statement is true otherwise enter '0'.

$\int a^{x+{\tan }^{-1}x}\left[\frac{x^2+2}{x^2+1}\right]dx$$=\frac{a^{x+{\tan }^{-1}\left(x\right)}}{\ln \left(a\right)}+c$

Answer 1

Let $I=\int a^{x+{\tan }^{-1}\left(x\right)}\frac{x^2+2}{x^2+1}dx$

substitute $u=x+{\tan }^{-1}\left(x\right)$

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

$I=\int a^{u}du$

$=\frac{a^u}{\ln \left(a\right)}$

$=\frac{a^{x+{\tan }^{-1}\left(x\right)}}{\ln \left(a\right)}+C$