Question

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

Answer 1

We have,

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

Let $t=x-4$

$dt=dx$

Therefore,

$I=\int \frac{2(t+4)+7}{t^2}dt$

$I=\int \frac{2t+8+7}{t^2}dt$

$I=\int \frac{2t+15}{t^2}dt$

$I=\int \frac{2t}{t^2}dt+\int \frac{15}{t^2}dt$

$I=\ln \left(t^2\right)-\frac{15}{t}+C$

On putting the value of $t$, we get

$I=\ln \left({(x-4)}^2\right)-\frac{15}{(x-4)}+C$

Hence, this is the answer.