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Q 22   Integrate the functions \frac{x + 3 }{x ^ 2 - 2x - 5 }

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Let (x+3) =A\frac{d}{dx}(x^2-2x+5)+B= A(2x-2)+B

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

\\\int \frac{x+3}{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\\ I=I_1+I_2..............(i)

\\\Rightarrow I_1\\ =\int \frac{2x-2}{x^2-2x-5}dx
put x^2-2x-5 =t \Rightarrow (2x-2)dx =dt

=\int \frac{dt}{t} = \log t = \log (x^2-2x-5)

\\\Rightarrow I_2\\ = \int \frac{1}{x^2-2x-5}dx\\ =\int \frac{1}{(x-1)^2+(\sqrt{6})^2}dx\\ =\frac{1}{2\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})

I=I_1+I_2

=\frac{1}{2}\log\left | x^2-2x-5 \right |+\frac{2}{\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})+C

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manish

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