Question

Evaluate the following $\underset{1}{\stackrel{2}{\int }}\frac{5x^2}{x^2+4x+3}$

Answer 1

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

$=5{\int }_1^{2}1+\frac{-4x-3}{x^2+4x+3}dx$

$=5{\int }_1^{2}1+\frac{-4x-3}{(x+2{)}^2-1}dx$

$u=x+2\to du=dx$

$=5{\int }_1^{2}1+\frac{-4u+5}{u^2-1}du$

$=5(u-2\ln |u^2-1|-5(\frac{1}{2}\ln |u+1|-\frac{1}{2}\ln |u-1|))$

$=5(x+2-2\ln \left|\right(x+2{)}^2-1|-5(\frac{1}{2}\ln |x+2+1|-\frac{1}{2}\ln |x+2-1|))$

$=5(x+2-2\ln |x^2+4x+3|-5(\frac{1}{2}\ln |x+3|-\frac{1}{2}\ln |x+1|))+C$

${\int }_1^2\frac{5x^2}{x^2+4x+3}dx=5(4-2\ln (15)-5\frac{\ln 5-\ln 3}{2})-5(3+5\frac{\ln 2}{2}-11\ln 2)$

$=5+55\ln 2-10\ln 15+\frac{-25\ln \frac{5}{3}-25\ln 2}{2}$