# How to use substitution to solve this integral? ∫_0^1dx/(sqrtx+5root(3)x)

Dec 2, 2017

${\int}_{0}^{1} \frac{6 {u}^{5} \cdot \mathrm{du}}{\sqrt{{u}^{6}} + 5 {\left({u}^{6}\right)}^{\frac{1}{3}}} = 137 - 750 L n \left(\frac{6}{5}\right)$

#### Explanation:

After using $u = {x}^{\frac{1}{6}}$, $x = {u}^{6}$ and $\mathrm{dx} = 6 {u}^{5} \cdot \mathrm{du}$ transforms, this integral became,

${\int}_{0}^{1} \frac{6 {u}^{5} \cdot \mathrm{du}}{\sqrt{{u}^{6}} + 5 {\left({u}^{6}\right)}^{\frac{1}{3}}}$

=${\int}_{0}^{1} \frac{6 {u}^{5} \cdot \mathrm{du}}{{u}^{3} + 5 {u}^{2}}$

=${\int}_{0}^{1} \frac{6 {u}^{3} \cdot \mathrm{du}}{u + 5}$

=${\int}_{0}^{1} \frac{\left(6 {u}^{3} + 30 {u}^{2} - 30 {u}^{2} - 150 u + 150 u + 750 - 750\right) \cdot \mathrm{du}}{u + 5}$

=${\int}_{0}^{1} \frac{\left[\left(6 {u}^{2} - 30 u + 150\right) \cdot \left(u + 5\right) - 750\right] \cdot \mathrm{du}}{u + 5}$

=${\int}_{0}^{1} \left(6 {u}^{2} - 30 u + 150\right) \cdot \mathrm{du}$-${\int}_{0}^{1} \frac{750 \cdot \mathrm{du}}{u + 5}$

=${\left[2 {u}^{3} - 15 {u}^{2} + 150 u\right]}_{0}^{1} - {\left[750 L n \left(u + 5\right)\right]}_{0}^{1}$

=$137 - 750 L n \left(\frac{6}{5}\right)$