Browse Questions

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

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)$\int\frac{dx}{\sqrt{x^2-a^2}}=log\mid x+\sqrt{x^2-a^2}\mid.$
Given:$I=\int \frac{x+2}{\sqrt{x^2-1}}dx.$

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

x+2=A(2x)+B.

Equating the coefficients,

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

B=2.

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

$\;\;\;=\int\frac{x}{\sqrt{x^2-1}}dx+\int\frac{2}{\sqrt{x^2-1}}dx$.

Let $x^2-1=t.$

On differentiating we get,

Therefore $2xdx=dt.\Rightarrow xdx=dt.$

On substituting we get,

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

On integrating we get,

$\;\;\;=\frac{1}{2}\big(\frac{t^\frac{1}{2}}{\frac{1}{2}}\big)+2log\mid x+\sqrt{x^2-1}\mid+c.$

$\;\;\;=\sqrt t+2log\mid x+\sqrt{x^2-1}\mid+c.$

Substituting for t we get,

$\;\;\;=\sqrt{x^2-1}+2log\mid x+\sqrt{x^2-1}\mid+c.$