Browse Questions

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

$\begin{array}{1 1} \sqrt{x^2+2x+3}+\log|(x+1)+\sqrt{x^2+2x+3}|+c \\ \sqrt{x^2+2x+3}-\log|(x+1)-\sqrt{x^2+2x+3}|+c \\ \sqrt{x^2+2x+3}-\log|(x+1)+\sqrt{x^2-2x-3}|+c \\ \sqrt{x^2-2x-3}-\log|(x+1)+\sqrt{x^2+2x+3}|+c \end{array}$

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)$\int\frac{dx}{\sqrt{x^2+a^2}}=log|x+\sqrt{x^2+a^2}|+c.$
Given:$I=\int \frac{x+2}{\sqrt{x^2+2x+3}}dx.$

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

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

Now equating the coefficients of x we get,

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

equating the coefficients we get,

1=2A+B$\Rightarrow B=1$.

Hence A=1/2 and B=1.

$I=\int\frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}dx.$

On separating the terms we can write,

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

If $x^2+2x+3=t.$

Therefore $(2x+2)dx=dt.$ and $\sqrt{x^2+2x+3}=\sqrt{(x+1)^2+(\sqrt 2)^2}.$

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

On integrating we get,

$\;\;\;=\frac{1}{2}\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+log|(x-1)+\sqrt{x^2+2x+3}|+c.$

$\;\;\;=\frac{1}{2}\big(\frac{\sqrt t}{1/2}\big)+log|(x+1)+\sqrt{x^2+2x+3}|+c.$

Substituting for t ,

Hence $\int\frac{x+2}{\sqrt{x^2+2x+3}}dx=\sqrt{x^2+2x+3}+log|(x+1)+\sqrt{x^2+2x+3}|+c.$