Question

Evaluate

$\int \frac{x-1}{x^2-4x-5}dx$

Answer 1

We have,

$\int \frac{x-1}{x^2-4x+5}dx$

Then,

$\int \frac{x-1}{x^2-4x-5}dx$

$\Rightarrow \int \frac{(x-1)}{x^2-(5-1)x-5}dx$

$\Rightarrow \int \frac{(x-1)}{x^2-5x+1x-5}dx$

$\Rightarrow \int \frac{(x-1)}{(x-5)(x+1)}dx$

$\Rightarrow \int \frac{x-1+1-1}{(x-5)(x+1)}dx$

$\Rightarrow \int \frac{x+1-2}{(x-5)(x+1)}dx$

$\Rightarrow \int \frac{(x+1)}{(x-5)(x+1)}dx+\int \frac{-2}{(x-5)(x+1)}dx$

$\Rightarrow \int \frac{1}{(x-5)}dx+\int \frac{-2}{(x-5)(x+1)}dx$

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

$\Rightarrow \int \frac{1}{(x-5)}dx-2\int \frac{1}{x^2-4x+4-9}dx$

$\Rightarrow \int \frac{1}{(x-5)}dx-2\int \frac{1}{{(x-2)}^2-3^2}dx$

On integrating and we get,

$\log (x-5)-2\times \frac{1}{2\times 3}\log \left(\frac{x-2-3}{x-2+3}\right)+C$

$\Rightarrow \log (x-5)-\frac{1}{3}\log \left(\frac{x-5}{x+1}\right)+C$

Hence, this is the answer.