# Verify the following $\large\int\frac{2x+3}{x^2+3x}dx=log \mid x^2+3x\mid+C$

Toolbox:
• Proper rational expression of the form $\large\frac{1}{(x+a)(x+b)}$ can be resolved into partial fractions as $\large\frac{A}{x+a}+\frac{B}{x+b}$
• $\large\int \frac{dx}{x}=log |x|+c$
Given $\large\int\frac{2x+3}{x^2+3x}dx$
$x^2+3x=x(x+3)$
Therefore $I=\large\frac{2x+3}{x(x+3)}dx$
$\large\frac{2x+3}{x(x+3)}$ can be resolved into partial fractions as
$\large\frac{2x+3}{x(x+3)}=\frac{A}{x}+\frac{B}{x+3}$
=>$2x+3=A(x+3)+Bx$
To find the value of A and B, let us first put x=-3
$-6+3=A(0)+B(-3)$
$-3=-3B=>B=1$
next put x=0
$=>3=A(0+3)+0$
$=>3A=3\; or\; A=1$
Hence $A=1\;and\;B=1$
Therefore $\large\frac{2x+3}{x(x+3)}=\frac{1}{x}+\frac{+1}{x+3}$
Now substituting this in I we get
$I=\large\int\frac{1}{x}dx+\int\frac{1}{x+3}dx$
on integrating we get,
$I=log x+log(x+3)+c$
$=log |x(x+3)|+c$
$=log |x^2+3x|+c$
Hence $LHS=RHS$ is verified