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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