# How do you integrate int-(8x^3)/(-2x^4+5)^5 using substitution?

Aug 5, 2016

$\frac{- 1}{4 {\left(- 2 {x}^{4} + 5\right)}^{4}} + C$

#### Explanation:

We have:

$\int \frac{- 8 {x}^{3}}{- 2 {x}^{4} + 5} ^ 5 \mathrm{dx}$

When choosing a good substitution for functions such as these, which have mainly terms that resemble polynomial functions (i.e., not trigonometric or exponential), it's good to find powers that are one apart, since the lower can often be the derivative of the higher.

Also, its good to write a fraction's denominator, if it's multiple terms, as a single term via substitution. Both of these pieces of advice apply here:

Let $u = - 2 {x}^{4} + 5$. The derivative of this is then $\mathrm{du} = - 8 {x}^{3} \mathrm{dx}$.

The substitution can now take place easily:

$\int \frac{- 8 {x}^{3}}{- 2 {x}^{4} + 5} ^ 5 \mathrm{dx} = \int \frac{1}{u} ^ 5 \mathrm{du} = \int {u}^{-} 5 \mathrm{du}$

This can be solved using typical integration rules:

$= - \frac{1}{4} {u}^{-} 4 = - \frac{1}{4 {u}^{4}} = \frac{- 1}{4 {\left(- 2 {x}^{4} + 5\right)}^{4}} + C$

Aug 5, 2016

$\int {u}^{- 5} \cdot d u = - \frac{1}{4 {\left(- 2 {x}^{4} + 5\right)}^{4}} + C$

#### Explanation:

int -(8x^3)/(-2x^4+5)^5 d x=?

$\text{Substitute "u=-2x^4+5" ; } d u = - 8 {x}^{3} \cdot d x$

$\int \frac{d u}{{u}^{5}} = \int {u}^{- 5} \cdot d u$

$\text{So ;} \int {u}^{- n} \cdot d u = \frac{1}{- n + 1} {u}^{- n + 1}$

$\int {u}^{- 5} \cdot d u = \frac{1}{- 5 + 1} {u}^{- 5 + 1}$

$\int {u}^{- 5} \cdot d u = - \frac{1}{4} {u}^{- 4}$

$\text{undo substitution}$

$\int {u}^{- 5} \cdot d u = - \frac{1}{4} {\left(- 2 {x}^{4} + 5\right)}^{- 4}$

$\int {u}^{- 5} \cdot d u = - \frac{1}{4} {\left(- 2 {x}^{4} + 5\right)}^{- 4} + C$

$\text{Or }$

$\int {u}^{- 5} \cdot d u = - \frac{1}{4 {\left(- 2 {x}^{4} + 5\right)}^{4}} + C$