# How do you integrate int x^3*(1-x^4)^6 dx ?

Apr 5, 2018

$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$

#### Explanation:

$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} - - - \left(1\right)$

method 1: inspection

we note that the function outside the bracket is a multiple of the bracket differentiated, so we can approach this integral by inspection.

We will 'guess' the bracket to the power +1, differentiate and compare it with the required integral.

$\frac{d}{\mathrm{dx}} {\left(1 - {x}^{4}\right)}^{7} = 7 {\left(1 - {x}^{4}\right)}^{6} \times \left(- 4 {x}^{3}\right)$

$= - 28 {x}^{3} {\left(1 - {x}^{4}\right)}^{6} - - \left(2\right)$

comparing $\left(1\right) \text{ & } \left(2\right)$ we can conclude

$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$

method 2 : substitution

$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} - - - \left(1\right)$

$u = 1 - {x}^{4} \implies \mathrm{du} = - 4 {x}^{3} \mathrm{dx}$

$\left(1\right) \rightarrow \int \cancel{{x}^{3}} {u}^{6} \times - \frac{1}{4 \cancel{{x}^{3}}} \mathrm{du}$

$= - \frac{1}{4} \int {u}^{6} \mathrm{du} = - \frac{1}{4} \times {u}^{7} / 7 + c$

$= - \frac{1}{28} {u}^{7} + c = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$