Question

Evaluate the definite integrals.
${\int }_1^2\frac{5x^2}{x^2+4x+3}dx$

Answer 1

Let $I={\int }_1^2\frac{5x^2}{x^2+4x+3}dx$
Here, the degree of numerator and denominator is same, so on dividing numerator by denominator, we get
$\frac{5x^2}{x^2+4x+3}=5-\frac{20x+15}{x^2+4x+3}\therefore I={\int }_1^2[5-\frac{20x+15}{x^2+4x+3}]dx={\int }_1^{2}5dx-{\int }_1^2\frac{20x+15}{(x+1)(x+3)}.....\left(i\right)$
Let $\frac{20x+15}{(x+1)(x+3)}dx=\frac{A}{(x+1)}+\frac{B}{(x+3)}$
$\Rightarrow 20x+15=A(x+3)+B(x+1)\Rightarrow 20x+15=Ax+3A+Bx+B$
On equating the coefficients of x and constant term on both sides, we get
20=A+B .....(ii)
and 15=3A +B .....(iii)
On solving Eqs.(ii)and (iii), we get
$A=-\frac{5}{2},B=\frac{45}{2}\therefore I=5{\int }_1^{2}1dx+\frac{5}{2}{\int }_1^2\frac{1}{(x+1)}dx-\frac{45}{2}{\int }_1^2\frac{1}{(x+3)}dx\left[fromEq\right(i\left)\right]={[5x+\frac{5}{2}log|x+1|-\frac{45}{2}log|x+3|]}_1^2=[10+\frac{5}{2}log|3|-\frac{45}{2}log|5|-5-\frac{5}{2}log|2|+\frac{45}{2}log|4|]\Rightarrow I=5+\frac{5}{2}log\left|\frac{3}{2}\right|-\frac{45}{2}log\left|\frac{5}{4}\right|[\because mloga-mlogb=mlog(\frac{a}{b}\left)\right]$