Question

Evaluate $\int \frac{x^3+{4x}^2-7x+5}{x+2}dx$

Answer 1

$\int \frac{x^3+4x^2-7x+5}{x+2}dx$

$u=x+2,du=dx$.

$x=u-2$

$x^2=(u-2{)}^2$=$u^2+4-4u$

$x^3=(u-2{)}^3$=$u^3-8-6u^2+12u$

$x^3+4x^2-7x+5=u^3-8-6u^2+12u+4u^2+16-16u-7u+14+5$

=$u^3-2u^2-11u+27$

=$\int \frac{u^3-2u^2-11u+27}{u}du$

=$\int (u^2-2u-11+\frac{27}{u})du$

=$\frac{u^3}{3}-u^2-11u+27\ln u+C$

=$\frac{(x+2{)}^3}{3}-(x+2{)}^2-11(x+2)+27|\ln x+2\mid +C$(Ans)