# How do you find the antiderivative of int 1/(3x-7)^2 dx from [3,4]?

Dec 6, 2016

We can rewrite this as

$\int {\left(3 x - 7\right)}^{-} 2 \mathrm{dx}$

We let $u = 3 x - 7$, then $\mathrm{du} = 3 \left(\mathrm{dx}\right)$, and $\mathrm{dx} = \frac{\mathrm{du}}{3}$.

$\implies {\int}_{3}^{4} {\left(u\right)}^{-} 2 \left(\frac{1}{3}\right) \mathrm{du}$

$\implies \frac{1}{3} {\int}_{3}^{4} {\left(u\right)}^{-} 2 \mathrm{du}$

$\implies \frac{1}{3} \left(- \frac{1}{2} {u}^{-} 1\right) {|}_{3}^{4}$

$\implies - \frac{1}{6 u} {|}_{3}^{4}$

$\implies - \frac{1}{6 \left(3 x - 7\right)} {|}_{3}^{4}$

$\implies - \frac{1}{18 x - 42} {|}_{3}^{4}$

We now evaluate using ${\int}_{a}^{b} \left(f \left(x\right)\right) = F \left(b\right) - F \left(a\right)$, where $F \left(x\right)$ is the antiderivative of $f \left(x\right)$.

$\implies - \frac{1}{18 \left(4\right) - 42} - \left(- \frac{1}{18 \left(3\right) - 42}\right)$

$\implies - \frac{1}{30} + \frac{1}{12}$

$\implies \frac{1}{20}$

Hopefully this helps!