Question

Integrate the function $\frac{x+3}{x^2-2x-5}$

Answer 1

Let $(x+3)=A\frac{d}{dx}(x^2-2x-5)+B$
$(x+3)=A(2x-2)+B$
Equating the coefficients of $x$ and constant term on both sides, we obtain
$2A=2\Rightarrow A=\frac{1}{2}$
& $-2A+B=3\Rightarrow B=4$
$\therefore (x+3)=\frac{1}{2}(2x-2)+4$
$\Rightarrow \int \frac{x+2}{x^2-2x-5}dx=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx$
$=\frac{1}{2}\int \frac{2x-2}{x^2-2x-5}dx+4\int \frac{1}{x^2-2x-5}dx$
Let $I_1=\int \frac{2x-2}{x^2-2x-5}dx$ and $I_2=\int \frac{1}{x^2-2x-5}dx$
$\therefore \int \frac{x+3}{(x^2-2x-5)}dx=\frac{1}{2}{I}_1+4I_2$ ...... (1)
Then, $I_1=\int \frac{2x-2}{x^2-2x-5}dx$
Let $x^2-2x-5=t$
$\Rightarrow (2x-2)dx=dt$
$\Rightarrow I_1=\int \frac{dt}{t}=\log |t|=\log |x^2-2x-5|$ .......... (2)
$I_2=\int \frac{1}{x^2-2x-5}dx$
$=\int \frac{1}{(x^2-2x-5)-6}dx$
$=\int \frac{1}{(x-1{)}^2+(\sqrt{6}{)}^2}dx$
$=\frac{1}{2\sqrt{6}}\log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)$ .........(3)
Substituting (2) and (3) in (1), we obtain
$\int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}\log |x^2-2x-5|+\frac{4}{2\sqrt{6}}\log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C$
$=\frac{1}{2}\log |x^2-2x-5|+\frac{2}{\sqrt{6}}\log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C$