Question

Solve :

$\int \frac{1}{{(7x-5)}^3}+\frac{1}{\sqrt{5x-4}}dx$

Answer 1

Given $\int \frac{1}{{(7x-5)}^3}+\frac{1}{\sqrt{5x-4}}dx$

$=\int {(7x-5)}^{-3}+\int {(5x-4)}^{\frac{-1}{2}}dx$

We know that $\int {(ax+b)}^{n}dx=\frac{1}{a(n+1)}{(ax+b)}^{n+1}$

$=\frac{1}{7(-3+1)}{(7x-5)}^{-3+1}+\frac{1}{5(\frac{-1}{2}+1)}{(5x-4)}^{\frac{-1}{2}+1}+C$

$=\frac{1}{7(-2)}{(7x-5)}^{-2}+\frac{1}{5\left(\frac{-1+2}{2}\right)}{(5x-4)}^{\frac{-1+2}{2}}+C$

$=\frac{1}{-14}{(7x-5)}^{-2}+\frac{2}{5}{(5x-4)}^{\frac{1}{2}}+C$