Question

Integration of root x over x plus 5? Thanks

Answer 1

`2(sqrtx-sqrt5tan^-1[sqrtx/sqrt5])+c`

Explanation:

Problem:
`intsqrtx/(x+5)dx`

Substitute `u->dx=2sqrtxdu`
Use:`x=u^2`
`2intu^2/(u^2+5)du`

Now solving:
`intu^2/(u^2+5)du`

Write `u^2` as `u^2+5-5` and split:
=`int((u^2+5)/(u^2+5)-5/(u^2+5))du`
=`int(1-5/(u^2+5))du`

Apply linearity:
=`int1du-5int1/(u^2+5)du`

Now solving:
`int1du`

Apply constant rule
=`u`

Now solving
=`int(1/u^2+5)du`

Substitute `v=u/sqrt5->du=sqrt5dv`
=`intsqrt5/(5v^2+5)dv`
=`1/sqrt5int1/(v^2+1)dv`

This is a standard integral:
`tan^-1v`

Plug in solved integrals:
`1/sqrt5int1/(v^2+1)dv`
=`tan^-1v/sqrt5`

Undo substitution `v=u/sqrt5`:
=`tan^-1(u/sqrt5)/sqrt5`

Plug in solved integrals:
`int1du-5int1/(u^2+5)du`
=`u-sqrt5tan^-1(u/sqrt5)`

Plug in solved integrals:
`2intu^2/(u^2+5)du`

Undo substitution `u=sqrtx`
=`2sqrtx-2sqrt5tan^-1(sqrtx/sqrt5)+c`

Simplify
=`2(sqrtx-sqrt5tan^-1[sqrtx/sqrt5])+c`