Question

How do you find the integral of `int x^3 * sqrt(x^2 + 4) dx`?

Answer 1

`1/15(x^2+4)^(3/2)*(3x^2-8)+C`

Explanation:

`int x^3sqrt(x^2+4)*dx`

After using `x=2tany` and `dx=2(secy)^2*dy` transforms, this integral became,

`int (2tany)^3sqrt((2tany)^2+4)*2(secy)^2*dy`

=`int 16(tany)^3*(secy)^2*sqrt(4(secy)^2)*dy`

=`int 32(secy)^3*(tany)^3*dy`

=`int 32(secy)^2*(tany)^2*secy*tany*dy`

=`int 32(secy)^2*((secy)^2-1)*secy*tany*dy`

After using `z=secy` and `dz=secy*tany*dy` transforms, it became

`int 32z^2*(z^2-1)*dz`

=`int (32z^4-32z^2)*dz`

=`32/5z^5-32/3z^3+C`

=`32/5(secy)^5-32/3(secy)^3+C`

After using `x=2tany`, `tany=x/2` and `secy=sqrt(x^2+4)/2` inverse transforms, I found

`1/5(x^2+4)^(5/2)-4/3(x^2+4)^(3/2)+C`

=`1/15(x^2+4)^(3/2)*(3x^2-8)+C`

Answer 2

`1/15(x^2+4)^(3/2)(3x^2-8)+C`.

Explanation:

Let, `I=intx^3sqrt(x^2+4)dx=intx^2sqrt(x^2+4)*xdx`.

Subst.

`x^2+4=t^2, or, x^2=t^2-4. :. 2xdx=2tdt, or, xdx=tdt`

`:. I=int(t^2-4)*sqrt(t^2)*tdt=int(t^2-4)t^2dt`,

`=int(t^4-4t^2)dt`,

`=t^5/5-4*t^3/3`,

`=t^3/15(3t^2-20)`,

`=1/15*t^2(3t^2-20)t`.

Reverting from `t^2 to (x^2+4)`, we have,

`I=1/15(x^2+4){3(x^2+4)-20}sqrt(x^2+4)`.

`rArr I=1/15(x^2+4)^(3/2)(3x^2-8)+C`, as Respected Cem

Sentin has readily derived.

Feel the Joy of Maths.!