Find the integral using trig substitution: x^3/sqrt(4+x^2)dx

Question

Calculus BC

Answer 1

desmos.com/calculator/5fq75y2mkx

Answer 2

integrate x^3/(x^2+4)^.5 dx

use u=x^2+4

du/dx= 2x

du=2xdx

dx= du/2x

=.5∫(u-4)/(u)^.5du

= .5∫(u^.5)-4u^-.5 du

=.5[2u^1.5/3 -8u^.5)] +C

= u^1.5/3 - 4u^.5 + C

replace u with x^2+4

= [(x^2 +4)^1.5]/3 - 4(x^2+4)^.5 + C

= [(x^2+4)^.5][(x^2+4)/3 -4] +C

= [sqr(x^2+4)(x^2+4)/3 -12/3)+ C

integral=[(x^2-8)[sqr(x^2+4)]/3 +C

you should get the same answer using trig sub, u-sub or any other method