# What is the improper integrals sqrtx ln 5x dx from 1 to e ?

May 11, 2018

$= \frac{2}{3} {e}^{\frac{3}{2}} \ln \left(5 e\right) - \frac{4}{9} {e}^{\frac{3}{2}} - \frac{2}{3} \ln \left(5\right) + \frac{4}{9}$

#### Explanation:

In preparation for evaluating the definite integral, we should first find the antiderivative $\int \sqrt{x} \ln \left(5 x\right) \mathrm{dx}$, which can be solved using Integration by Parts:

$u = \ln \left(5 x\right)$
$\mathrm{du} = {x}^{-} 1 \mathrm{dx}$
$\mathrm{dv} = \sqrt{x} \mathrm{dx}$
$v = \frac{2}{3} {x}^{\frac{3}{2}}$

$u v - \int v \mathrm{du} = \frac{2}{3} {x}^{\frac{3}{2}} \ln \left(5 x\right) - \frac{2}{3} \int {x}^{\frac{3}{2}} {x}^{- 1} \mathrm{dx}$

=2/3x^(3/2)ln(5x)-2/3intsqrtxdx

$= \frac{2}{3} {x}^{\frac{3}{2}} \ln \left(5 x\right) - \frac{4}{9} {x}^{\frac{3}{2}}$ (leaving out the constant as we're going to use this to evaluate a definite integral)

Now, we may evaluate the improper definite integral:

${\int}_{1}^{e} \sqrt{x} \ln \left(5 x\right) \mathrm{dx}$

This is not an improper integral; the integrand $\sqrt{x} \ln \left(5 x\right)$ is continuous on the interval of integration $\left[1 , e\right]$.

Thus,

${\int}_{1}^{e} \sqrt{x} \ln \left(5 x\right) \mathrm{dx} = \left[\frac{2}{3} {x}^{\frac{3}{2}} \ln \left(5 x\right) - \frac{4}{9} {x}^{\frac{3}{2}}\right] {|}_{1}^{e}$

$= \frac{2}{3} {e}^{\frac{3}{2}} \ln \left(5 e\right) - \frac{4}{9} {e}^{\frac{3}{2}} - \frac{2}{3} \ln \left(5\right) + \frac{4}{9}$