IMathAS-libretexts

Evaluate using the substitution method.

$\displaystyle \int{x}^{{2}}{\left({x}^{{3}}-{5}\right)}^{{8}}{\left.{d}{x}\right.}$

$\displaystyle \int{\left(\frac{{u}^{{2}}}{{2}}{d}{u}\right)}=\frac{{\left({x}^{{3}}-{5}\right)}^{{9}}}{{18}}+{C}$
$\displaystyle \int{\left(\frac{{u}^{{2}}}{{3}}{d}{u}\right)}=\frac{{x}^{{3}}}{{{27}}}\cdot{\left({x}^{{3}}-{5}\right)}^{{9}}\cdot{\left(\frac{{x}^{{4}}}{{4}}-{5}{x}\right)}+{C}$
$\displaystyle \int{\left(\frac{{u}^{{8}}}{{3}}{d}{u}\right)}=\frac{{x}^{{3}}}{{{27}}}\cdot{\left({x}^{{3}}-{5}\right)}^{{9}}\cdot{\left(\frac{{x}^{{4}}}{{4}}-{5}{x}\right)}+{C}$
$\displaystyle \int{\left(\frac{{u}^{{8}}}{{4}}{d}{u}\right)}=\frac{{\left({x}^{{3}}-{5}\right)}^{{9}}}{{27}}+{C}$
$\displaystyle \int{\left(\frac{{u}^{{9}}}{{27}}{d}{u}\right)}=\frac{{x}^{{3}}}{{{27}}}\cdot{\left(\frac{{x}^{{4}}}{{4}}-{5}{x}\right)}^{{9}}+{C}$
$\displaystyle \int{\left(\frac{{u}^{{8}}}{{3}}{d}{u}\right)}=\frac{{\left({x}^{{3}}-{5}\right)}^{{9}}}{{27}}+{C}$