# Integrate the functions$\frac{4x+1}{\sqrt{2x^2+x-3}}$

Toolbox:
• $\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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.
Given:$\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx.$

Let 4x+1=A.$\frac{d}{dx}(2x^2+x-3)+B.$

$\Rightarrow 4x+1=A(4x+1)+B.$

$\Rightarrow 4x+1=4Ax+A+B.$

On equating the coefficients we get,

4A=4$\Rightarrow A=1.$

A+B=1$\Rightarrow B=0.$

On substituting for A and B we get

Therefore $4x+1=1(4x+1)+0.$

Hence $I=\int\frac{4x+1}{\sqrt{2x^2+x-3}}$.

Let $2x^2+x-3=t.$

On differentiating we get,

Therefore $(4x+1)dx=dt.$

On substituting we get,

$I=\int\frac{dt}{\sqrt t}.$

On integrating we get,

$\;\;\;=\frac{t^\frac{1}{2}}{\frac{1}{2}}+c.$

$\;\;\;=2\sqrt t+c.$

Substituting for t we get,

$\;\;\;=2\sqrt{2x^2+x-3}+c.$

answered Feb 4, 2013