Question

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

Answer 1

Let $I={\int }_1^2\frac{5x^2}{x^2+4x+3}$
Since, degree of numerator is equal to degree of denominator, so dividing $5x^2$ by $x^2+4x+3$

$I={\int }_1^2\{5-\frac{20x+15}{x^2+4x+3}\}dx$

$={\int }_1^{2}5dx-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

$=[5x{]}_1^2-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

$I=5-I_1$ ........... (1)

where $I_1={\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

Consider $I_1={\int }_1^2\frac{20x+15}{x^2+4x+3}dx$
Let $20x+15=A\frac{d}{dx}(x^2+4x+3)+B$

$\Rightarrow 20x+15=A(2x+4)+B$ ....(2)
$=2Ax+(4A+B)$
Equating the coefficients of $x$ and constant term, we obtain
$A=10$ and $B=-25$

Substituting these values in (2),

$20x+15=10(2x+4)-25$
$\Rightarrow I_1=10{\int }_1^2\frac{2x+4}{x^2+4x+3}dx-25{\int }_1^2\frac{dx}{x^2+4x+3}$
Let $x^2+4x+3=t$
$\Rightarrow x^2+4x+3=t$
$\Rightarrow (2x+4)dx=dt$

$\Rightarrow I_1=10{\int }_8^{15}\frac{dt}{t}-25{\int }_1^2\frac{dx}{(x+2{)}^2-1^2}$

$=10[\log t{]}_8^{15}-25{\left[\frac{1}{2}\log \left(\frac{x+2-1}{x+2+1}\right)\right]}_1^2$

$=10[\log 15-\log 8]-25{\left[\frac{1}{2}\log \left(\frac{x+1}{x+3}\right)\right]}_1^2$

$=[10\log 15-10\log 8]-25[\frac{1}{2}\log \frac{3}{5}-\frac{1}{2}\log \frac{2}{4}]$

$=\left[10\log \right(5\times 3)-10\log (4\times 2\left)\right]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10\log 5+10\log 3-10\log 4-10\log 2-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10+\frac{25}{2}]\log 5+[-10-\frac{25}{2}]\log 4+[10-\frac{25}{2}]\log 3+[-10+\frac{25}{2}]\log 2$

$=\frac{45}{2}\log 5-\frac{45}{2}\log 4-\frac{5}{2}\log 3+\frac{5}{2}\log 2$

$=\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Substituting the value of $I_1$ in (1), we get
$I=5-[\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}]$

$=5-\frac{5}{2}[9\log \frac{5}{4}-\log \frac{3}{2}]$