Evaluate: `int(x-5)sqrt(x^2+x )dx`

Explanation

Let ` (x-5)=lambda (d)/(dx)(x^(2)+x) +mu`
Then, `x-5=lambda(2x+1)+mu`
Comparing cofficients of like power of x, we get
` 1=2lambda " and " mu= -5`
or `lambda=(1)/(2) " and " mu =(11)/(2). ` Therefore,
`int (x-5)sqrt(x^(2)+x)dx=int[(1)/(2)(2x+1)-(11)/(2)]sqrt(x^(2)+x)dx`
`=(1)/(2)int(2x+1)sqrt(x^(2)+x)dx-(11)/(2)intsqrt(x^(2)+x)dx`
`=(1)/(2)intsqrt(t)dt-(11)/(2)intsqrt((x+(1)/(2))^(2)-((1)/(2))^(2))dx, " where " t=x^(2)+x`
`=(1)/(2)(t^(3//2))/(3//2)-(11)/(2)[(1)/(2)(x+(1)/(2))sqrt((x+(1)/(2))^(2)-((1)/(2))^(2))-(1)/(2)((1)/(2))^(2) log |(x+(1)/(2))+sqrt((x+(1)/(2))^(2)-((1)/(2))^(2))|]+C`
`=((x^(2)+x)^(3//2))/(3)-(11)/(2)[(2x+1)/(4)sqrt(x^(2)+x)-(1)/(8)log|(x+(1)/(2))+sqrt(x^(2)+x)|]+C`