Let $I=\int \frac{x+3}{x^2-2x-5} dx$ Put $x+3=A\left(\frac{d}{dx}(x^2-2x-5)\right)+B$ $=A(2x-2)+B$ ... (i) Comparing the coefficient of x in (i), we get $1 = 2A \Rightarrow A=\frac{1}{2}$ Comparing the constant terms in (i), we get $3 = B - 2A$ $\Rightarrow 3 = B - 1$ $\Rightarrow B = 4$ $I=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5} dx$ $=\frac{1}{2}\int \frac{2x-2}{x^2-2x-5} dx+4\int \frac{dx}{x^2-2x-5}$ Let $I=\frac{1}{2}I_1+4I_2$ ... (ii) where $I_1=\int \frac{2x-2}{x^2-2x-5} dx$ Put $x^2 - 2x - 5 = t$ $\Rightarrow (2x - 2) dx = dt$ $\therefore I_1=\int \frac{dt}{t}=\log{|t|}$ $=\log{| x^2 - 2x - 5 |}+C_1$ ... (iii) and $I_2=\int \frac{dx}{x^2 - 2x - 5 } =\int \frac{dx}{ (x-1)^2 - 6 }$ $=\int \frac{dx}{(x-1)^2 - (\sqrt{6})^2 }$ $=\frac{1}{2\sqrt{6}} \log{\left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|}+C_2$ .. (iv) Hence from (ii), (iii) and (iv), we get