# Integrate the functions$\frac{5x+3}{\sqrt{x^2+4x+10}}$

$\begin{array}{1 1} 5\log|x^2+4x+10|-7\log|(x+2)+\sqrt{x^2+4x+10}|+c \\5\log|x^2+4x+10|+7\log|(x+2)+\sqrt{x^2+4x+10}|+c \\ 5\log|x^2+4x+10|+7\log|(x+2)-\sqrt{x^2+4x+10}|+c \\ 5\log|x^2+4x+10|-7\log|(x+2)-\sqrt{x^2+4x+10}|+c \end{array}$

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)$\int\frac{dx}{\sqrt{x^2+a^2}}=log\mid x+\sqrt{x^2+a^2}\mid+c.$
Given:$I=\int \frac{5x+3}{x^2+4x+10}dx.$

Let $5x+3=A\frac{d}{dx}(x^2+4x+10)+B.$

$\quad 5x+3=A(2x+4)+B.$

Now equating the coefficients of x we get,

5=2A$\Rightarrow A=\frac{5}{2}.$

equating the coefficients we get,

3=4A+B$\Rightarrow B=-7\;and\;A=\frac{5}{2}.$

Hence A=5/2 and B=-7.

$I=\int\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx.$

On separating the terms we can write,

$I=\frac{5}{2}\int\frac{2x+42}{\sqrt{x^2+4x+10}}-7\int\frac{dx}{\sqrt{x^2+4x+10}}.$

Let $x^2+4x+10=t.$

Therefore (2x+4)dx=dt.

And $x^2+4x+10$ can be written as $(x+2)^2-(\sqrt 6)^2.$

On substituting we get,

$I=\frac{5}{2}\int\frac{dt}{t}-7\int\frac{dx}{(x+2)^2+(\sqrt 6)^2}.$

On integrating we get,

$\;\;\;=\frac{5}{2}\begin{bmatrix}\frac{t^{-1/2+1}}{-1/2+1}\end{bmatrix}-7log \mid(x+2)+\sqrt{x^2+4x+10}\mid+c.$

Substituting for t we get,

$\;\;\;=5log|x^2+4x+10|-7log|(x+2)+\sqrt{x^2+4x+10}|+c.$