# How do you find the antiderivative of int 1/root3(1-5t) dt?

May 23, 2017

I got: $- \frac{3}{10} \sqrt[3]{{\left(1 - 5 t\right)}^{2}} + c$

#### Explanation:

Have a look:

May 24, 2017

$\int \left(\frac{1}{\sqrt[3]{1 - 5 t}}\right) \mathrm{dt} = - \frac{3}{10} {\left(1 - 5 t\right)}^{\frac{2}{3}} + C$

#### Explanation:

$\int \left(\frac{1}{\sqrt[3]{1 - 5 t}}\right) \mathrm{dt}$

rewrite as

$\int {\left(1 - 5 t\right)}^{- \frac{1}{3}} \mathrm{dt}$

now the outside of the bracket $= \text{ constant "xx " bracket differentiated}$

so we are able to do this by inspection

using the power rule for integration ( add one to the power), and the fact that integration is the reverse of differentiation, let us try

$\frac{d}{\mathrm{dx}} {\left(1 - 5 t\right)}^{\frac{2}{3}}$

by the chain rule we get

$= \frac{2}{3} \times \left(- 5\right) {\left(1 - 5 t\right)}^{- \frac{1}{3}} = - \frac{10}{3} {\left(1 - 5 t\right)}^{- \frac{1}{3}}$

so by comparing the integral and our 'inspected solution' we conclude

$\int \left(\frac{1}{\sqrt[3]{1 - 5 t}}\right) \mathrm{dt} = - \frac{3}{10} {\left(1 - 5 t\right)}^{\frac{2}{3}} + C$