Question

If the integral $I=\displaystyle \int e^{5 ln x}\left(x^{6} + 1\right)^{- 1}dx$ $=\lambda ln\left(x^{6} + 1\right)+C$ , (where, $C$ is the constant of integration) then the value of $\frac{1}{\lambda }$ is

Answer 1

The integral
$I=\displaystyle \int e^{5 ln x}\left(x^{6} + 1\right)^{- 1}dx$
$I=\displaystyle \int \frac{e^{ln \left(x^{5}\right)}}{x^{6} + 1}dx$
$=\displaystyle \int \frac{x^{5}}{x^{6} + 1}dx$
$=\frac{1}{6}\displaystyle \int \frac{6 x^{5} d x}{x^{6} + 1}$
Put $x^{6}+1=t$
$6x^{5}dx=dt$
$\Rightarrow I=\frac{1}{6}\displaystyle \int \frac{d t}{t}=\frac{1}{6}ln\left(t\right)+C$
$\Rightarrow I=\frac{1}{6}ln\left(x^{6} + 1\right)+C$
$\Rightarrow \lambda =\frac{1}{6}$
$\Rightarrow \frac{1}{\lambda }=6$