Question

Don't try to spam here otherwise report​​

Answer 1

$\displaystyle{ \bold \red{\int{\frac{x^{3}}{a + x} d x}}}$

${ \displaystyle \bold \red{\color{red}{\int{\frac{x^{3}}{a + x} d x}} = \color{red}{\int{\left(- \frac{a^{3}}{a + x} + a^{2} - a x + x^{2}\right)d x}}}}$

$\displaystyle \bold {\color{red}{\int{\left(- \frac{a^{3}}{a + x} + a^{2} - a x + x^{2}\right)d x}} = \color{red}{\left(\int{a^{2} d x} + \int{x^{2} d x} - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x}\right)}}$

$\displaystyle \bold \red{{\int{x^{2} d x} - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x} + \color{red}{\int{a^{2} d x}} = \int{x^{2} d x} - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x} + \color{red}{a^{2} x}}}$

$\displaystyle {\bold \red{a^{2} x - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x} + \color{red}{\int{x^{2} d x}}=a^{2} x - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x} + \color{red}{\frac{x^{1 + 2}}{1 + 2}}=a^{2} x - \int{a x d x} - \int{\frac{a^{3}}{a + x} d x} + \color{red}{\left(\frac{x^{3}}{3}\right)}}}$

$\displaystyle{ \bold \red{a^{2} x + \frac{x^{3}}{3} - \int{\frac{a^{3}}{a + x} d x} - \color{red}{\int{a x d x}} = a^{2} x + \frac{x^{3}}{3} - \int{\frac{a^{3}}{a + x} d x} - \color{red}{a \int{x d x}}}}$

${\displaystyle \bold \red{a^{2} x - a \color{red}{\int{x d x}} + \frac{x^{3}}{3} - \int{\frac{a^{3}}{a + x} d x}=a^{2} x - a \color{red}{\frac{x^{1 + 1}}{1 + 1}} + \frac{x^{3}}{3} - \int{\frac{a^{3}}{a + x} d x}=a^{2} x - a \color{red}{\left(\frac{x^{2}}{2}\right)} + \frac{x^{3}}{3} - \int{\frac{a^{3}}{a + x} d x}}}$

${\displaystyle \bold \red{a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3} - \color{red}{\int{\frac{a^{3}}{a + x} d x}} = a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3} - \color{red}{a^{3} \int{\frac{1}{a + x} d x}}}}$

${ \displaystyle \bold \red{- a^{3} \color{red}{\int{\frac{1}{a + x} d x}} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3} = - a^{3} \color{red}{\int{\frac{1}{u} d u}} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3}}}$

$\displaystyle{ \bold \red{- a^{3} \color{red}{\int{\frac{1}{u} d u}} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3} = - a^{3} \color{red}{\ln{\left(u \right)}} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3}}}$

${ \displaystyle \bold \red{- a^{3} \ln{\left(\color{red}{u} \right)} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3} = - a^{3} \ln{\left(\color{red}{\left(a + x\right)} \right)} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3}}}$

Therefore,

${\displaystyle \bold \red{\int{\frac{x^{3}}{a + x} d x} = - a^{3} \ln{\left(\left|{a + x}\right| \right)} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3}}}$

${ \displaystyle \bold \red{ \int{\frac{x^{3}}{a + x} d x} = - a^{3} \ln{\left(\left|{a + x}\right| \right)} + a^{2} x - \frac{a x^{2}}{2} + \frac{x^{3}}{3}+C}}$