Question

Integrate the following functions.
$\int \frac{x+3}{x^2-2x-5}dx.$

Answer 1

Let $(x+3)=A\frac{d}{dx}(x^2-2x-5)+B$
$\Rightarrow (x+3)=A(2x-2)+B\Rightarrow x+3=2Ax-2A+B$
On equating the coefficients of x and constant term on both sides, we get
$2A=1\Rightarrow A=\frac{1}{2}and-2A+B=3\Rightarrow B=4\Rightarrow (x+3)=\frac{1}{2}(2x-2)+4\therefore \int \frac{x+3}{x^2-2x-5}dx=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx=\frac{1}{2}\int \frac{(2x-2)}{x^2-2x-5}dx+4\int \frac{1}{x^2-2x-5}dx$
Let $I_1=\int \frac{2x-2}{x^2-2x-5}dxand{I}_2\int \frac{1}{x^2-2x-5}dx$
$\therefore \int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}{I}_1+4I_2.......\left(i\right)$
Now, $I_1=\int \frac{2x-2}{x^2-2x-5}dx$
Let $x^2-2x-5=t\Rightarrow (2x-2)dx=dt$
$\therefore I_1=\int \frac{dt}{t}=log|t|+C_1=log|x^2-2x-5|+C_1.......\left(ii\right)$
and $I_2=\int \frac{1}{x^2-2x-5}dx=\int \frac{1}{(x^2-2x+1)-6}dx=\int \frac{1}{(x-1{)}^2-(\sqrt{6}{)}^2}dx$
$=\frac{1}{2\sqrt{6}}log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|.........\left(iii\right)[\because \int \frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|]$
On substituting the values $I_1$ and $I_2$ from Eqs. (ii) and (iii) in Eq (i), we get

$\int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}log|x^2-2x-5|+4\left[\frac{1}{2\sqrt{6}}log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\right]+C[\therefore \frac{1}{2}{C}_1+4C_2=C]=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt{6}}log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C$