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

$\begin{array}{1 1} \frac{1}{2}\log|x^2-2x-5|+\frac{2}{\sqrt 6}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|-\large \frac{2}{\sqrt 6}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|+\large \frac{1}{\sqrt 3}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \\ \frac{1}{2}\log|x^2-2x-5|- \frac{2}{\sqrt 3}\log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c \end{array}$

Toolbox:
• $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,where p,q,a,b,c are constants,we are to find real numbers such that $px+q=A\frac{d}{dx}(ax^2+bx+c)+B.$
• Therefore $px+q=A(2ax+b)+B.$
• To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms:
• (ii)$\int\frac{dx}{\sqrt{x^2-a^2}}=\frac{1}{2a}log\mid\frac{x-a}{x+a}\mid+c.$
Given:$I=\int \frac{x+3}{x^2-2x-5}dx.$

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

$\;\;\;\;\quad =A(2x-2)+B.$

Now equating the coefficients of x we get,

1=2A$\Rightarrow A=\frac{1}{2}.$

equating the coefficients we get,

3=-2A+B$\Rightarrow B=4\;and\;A=\frac{1}{2}.$

Hence A=1/2 and B=4.

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

On separating the terms we can write,

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

Let $x^2-2x-5=t.$

Therefore (2x-2)dx=dt.

And $x^2-2x-5$ can be written as $(x-1^2-(\sqrt 6)^2.$

On substituting we get,

$I=\frac{1}{2}\int\frac{dt}{t}+4\int\frac{dx}{(x-1)^2-(\sqrt 6)^2}.$

On integrating we get,

$\;\;\;=\frac{1}{2}log t+4.\frac{1}{2\sqrt 6}log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c.$+c.$Substituting for t , Hence$\int\frac{x+3}{x^2-2x-5}dx=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt 6}log\mid\frac{x-1-\sqrt 6}{x-1+\sqrt 6}\mid+c.\$