Question

Evaluate: ${\int }_1^2\frac{5x^2}{x^2+4x+3}dx$

Answer 1

${\int }_1^2\frac{5x^{2}dx}{x^2+4x+3}$

$={\int }_1^2\frac{5x^{2}dx}{x^2+x+3x+3}$

$={\int }_1^2\frac{5x^{2}dx}{x(x+1)+3(x+1)}$

$={\int }_1^2\frac{5x^{2}dx}{(x+1)(x+3)}$

Consider $\frac{5x^2}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$

$\Rightarrow 5x^2=A(x+3)+B(x+1)$

Put $x=-1\Rightarrow 5=2A\therefore A=\frac{5}{2}$

Put $x=-3\Rightarrow 45=-2B\therefore B=\frac{-45}{2}$

Now,${\int }_1^2\frac{5x^{2}dx}{x^2+4x+3}$

$=\frac{5}{2}{\int }_1^2\frac{dx}{x+1}-\frac{45}{2}{\int }_1^2\frac{dx}{x+3}$

$=\frac{5}{2}{\left[\log |x+1|\right]}_1^2-\frac{45}{2}{\left[\log |x+3|\right]}_1^2$

$=\frac{5}{2}[\log |2+1|-\log |1+1|]-\frac{45}{2}[\log |2+3|-\log |1+3|]$

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

$=\frac{5}{2}\log \left|\frac{3}{2}\right|+\frac{45}{2}\log \left|\frac{4}{5}\right|$