Question

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

Answer 1

Answer: 6

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