Question

Solve ${\int }_0^4\frac{x^2}{1+x}dx$

Answer 1

${\int }_0^4\frac{x^2}{x+1}dx$

Substitute $u=x+1$

$\Rightarrow dx=du$ & $x^2=(u-1{)}^2$

$={\int }_0^4\frac{(u-1{)}^2}{u}dx$

$={\int }_0^4(u+\frac{1}{u}-2)du$

$={\int }_0^{4}udu+{\int }_0^4\frac{1}{u}du-{\int }_0^{2}2du$

$={\left[\frac{u^2}{2}\right]}_0^4+{\left[lnu\right]}_0^4-2[u{]}_0^4$

$={\left[\frac{(x+1{)}^2}{2}\right]}_0^4+{\left[ln\right(x+1\left)\right]}_0^4-2[x+1{]}^4$

$=[\frac{25}{2}-\frac{1}{2}]+\left[ln\right(5)-ln(1\left)\right]-2[5-1]$

$=\left[\frac{24}{2}\right]+ln\left(5\right)-8$

$=ln\left(5\right)+4$

$=4,6989$.