Question

Solve the integral:

$\displaystyle \rm\int\dfrac{x^{\frac{x}{ln\:x}}}{\sqrt{e^x}} dx$

Answer 1

Given integral,

$\longrightarrow \small{\displaystyle \rm\int\dfrac{x^{\frac{x}{ln\:x}}}{\sqrt{e^x}} dx}$

We know that,

• $\boxed{\rm e^{\ln x} = x}$

By using this, we get:

$\longrightarrow \small{\displaystyle \rm\int\dfrac{ ({e}^{ \ln x}{ )^{\frac{x}{ln\:x}}}}{\sqrt{e^x}} dx}$

$\longrightarrow \small{\displaystyle \rm\int\dfrac{ {e}^{x} }{\sqrt{e^x}} dx}$

$\longrightarrow \small{\displaystyle \rm\int\dfrac{ {e}^{x} }{e^ \frac{x}{2} } dx}$

$\longrightarrow \small{\rm\int{e}^{x - \frac{x}{2} }dx}$

$\longrightarrow \small{\rm\int{e}^{ \frac{x}{2} }dx}$

Now, substitute $\rm u = \dfrac{x}{2} \longrightarrow \dfrac{du}{dx} = \dfrac{1}{2}$.

So,

$\longrightarrow \small{\rm\int{e}^{ \frac{x}{2} }dx \implies \int {e}^{u} 2 \: du}$

$\longrightarrow \small{\rm2 \int {e}^{u} \: du}$

$\longrightarrow \small{\rm2 {e}^{u} + C}$

Substitute back $\rm u = \dfrac{x}{2}$

$\longrightarrow \small{\rm2 {e}^{ \frac{x}{2} } + C}$

$\longrightarrow \small{\rm 2 \sqrt{ {e}^{x} } +C}$

Therefore,

$\underline{ \boxed{ \small{\displaystyle \rm\int\dfrac{x^{\frac{x}{ln\:x}}}{\sqrt{e^x}} dx= 2 \sqrt{ {e}^{x} }+C}}}$

Note : $\rm\int e^u du = e^u$