# How do you intergrate \frac{ (\sqrt{ln \ r})}{r}

## Question:

How do you integrate {eq}\frac{ (\sqrt{ln \ r})}{r} {/eq}

## Integration By Substitution:

There are many ways to solve the integral but integration by substitution is one of the best methods to solve the difficult integral.

All integrals are not solved by using the substitution method. Only those integrals are solved by this method which has the form {eq}\int f(g(x))g'(x) dx {/eq}.

The substitution method is a technique of converting one form of integral to another form integral by changing the variable and after solving the integration we reverse the substitution to get the answer back in terms of the given variable.

We have to integrate the given function

{eq}\int \dfrac{ (\sqrt{ln \ r})}{r} dr {/eq}

We use the substitution {eq}\ln (r)=t {/eq} to solve the integral.

Differentiate both sides of the substitution with respect to {eq}r {/eq}

{eq}\begin{align} \ln (r) &=t\\ \dfrac{\mathrm{d} }{\mathrm{d} r}(\ln (r)) &=\dfrac{\mathrm{d} t}{\mathrm{d} r}\\ \dfrac{1}{r} &=\dfrac{\mathrm{d} t}{\mathrm{d} r}\\ \dfrac{1}{r}dr &=dt \end{align} {/eq}

Applying this substitution to the integral, we have:

{eq}\begin{align} \int \dfrac{ (\sqrt{ln \ r})}{r} dr &=\int \sqrt{t}dt\\ &=\int t^{\dfrac{1}{2}}dt\\ &=\dfrac{t^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1}+C\\ &=\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}+C\\ &=\dfrac{2}{3}t^{\dfrac{3}{2}}+C \end{align} {/eq}

Reversing the substitution, we have:

{eq}\begin{align} \int \dfrac{ (\sqrt{ln \ r})}{r} dr &=\dfrac{2}{3}t^{\dfrac{3}{2}}+C\\ &=\dfrac{2}{3}\left ( \ln (r) \right )^{\dfrac{3}{2}}+C \end{align} {/eq}

{eq}\color{blue}{\int \dfrac{ (\sqrt{ln \ r})}{r} dr =\dfrac{2}{3}\left ( \ln (r) \right )^{\dfrac{3}{2}}+C} {/eq}