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

2 Answers
May 29, 2018

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

May 29, 2018

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.

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