Question

Integral:`int(12x^3)/sqrt(2x^2+7)dx` ?

`int(12x^3)/sqrt(2x^2+7)dx`

Answer 1

`sqrt(2x^2+7) (2x^2-14)+C`

Explanation:

Substitute `2x^2+7=u^2` so that `4xdx = 2udu` and `x^2=(u^2-7)/2`

Thus

`int(12x^3)/sqrt(2x^2+7)dx=int(6x^2)/sqrt(2x^2+7) 2xdx`
`qquad = int {3(u^2-7)}/{u} udu = 3int (u^2-7)du`
`qquad = u^3-21u+C = u(u^2-21)+C`
`qquad =sqrt(2x^2+7) (2x^2-14)+C`

Answer 2

`I=2sqrt(2x^2+7)(x^2-7)+c`

Explanation:

Here,

`I=int(12x^3)/sqrt(2x^2+7)dx`

`=int(12x^2)/sqrt(2x^2+7)*xdx`

Let, `2x^2+7=u^2=>2x^2=u^2-7=>4xdx=2udu`

`=>xdx=1/2udu`

So,

`I=int(12(u^2-7)/2)/canceluxx1/2canceludu`

`=3int(u^2-7)du`

`=3[u^3/3-7u]+c`

`=u^3-21u+c...to` where , `2x^2+7=u^2`

`=u(u^2-21)+c`

`=sqrt(2x^2+7)(2x^2+7-21)+c`

`=sqrt(2x^2+7)(2x^2-14)+c`

`=2sqrt(2x^2+7)(x^2-7)+c`