# How do you use Integration by Substitution to find intdx/(1-6x)^4dx?

Aug 6, 2014

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

We will let $u = 1 - 6 x$. Thus, $\mathrm{du} = - 6 \mathrm{dx}$.

$= \int {u}^{- 4} \mathrm{dx}$

This looks difficult since there isn't a $- 6$ in there for us to form a $\mathrm{du}$ with. However, there is a rule of integration which states:

$\int c \cdot f \left(x\right) \mathrm{dx} = c \cdot \int f \left(x\right) \mathrm{dx}$

We can exploit this rule to rewrite our integral equivalently as:

$= - \frac{1}{6} \int - 6 {u}^{- 4} \mathrm{dx}$

The statements are completely equivalent; note that if we pull the $- 6$ out of the integral, then it cancels with the $- \frac{1}{6}$ and leaves us with $1$ multiplied by our original integral.

Anyway, we now have a $- 6 \mathrm{dx}$ to form a $\mathrm{du}$ with.

$= - \frac{1}{6} \int {u}^{- 4} \mathrm{du}$

$= - \frac{1}{6} {u}^{- 3} \cdot \left(- \frac{1}{3}\right)$

$= \frac{1}{18} {u}^{- 3}$

$= \frac{1}{18 {u}^{3}}$

Substituting back for $u$ gives us:

$= \frac{1}{18 {\left(1 - 6 x\right)}^{3}}$