# How do you integrate int_1^7 sqrtx*lnx  using integration by parts?

Dec 4, 2016

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} \sqrt{343} \left(\ln 7 - \frac{2}{3}\right) + \frac{4}{9}$

#### Explanation:

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} {\int}_{1}^{7} \ln x d \left({x}^{\frac{3}{2}}\right)$

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} \left({x}^{\frac{3}{2}} \ln x\right) {|}_{x = 1}^{x = 7} - \frac{2}{3} {\int}_{1}^{7} {x}^{\frac{3}{2}} d \left(\ln x\right)$

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} \left({x}^{\frac{3}{2}} \ln x\right) {|}_{x = 1}^{x = 7} - \frac{2}{3} {\int}_{1}^{7} {x}^{\frac{3}{2}} \frac{\mathrm{dx}}{x}$

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} \left({x}^{\frac{3}{2}} \ln x\right) {|}_{x = 1}^{x = 7} - \frac{2}{3} {\int}_{1}^{7} {x}^{\frac{1}{2}} \mathrm{dx}$

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} \left(\ln x - \frac{2}{3}\right) {|}_{x = 1}^{x = 7}$

${\int}_{1}^{7} \sqrt{x} \ln x \mathrm{dx} = \frac{2}{3} \sqrt{343} \left(\ln 7 - \frac{2}{3}\right) + \frac{4}{9}$