what is the integral of (x^3)/(sqr(x^2+9)) dx?

Answer 1

you can anwer this by subtituation
put u^2 = ((x^2+9))
derivate it 2u du/dx = 2x +0
dx =2u du/dx

so, int {x^3/u 2udu/2x} cancle out
u get int {x^2 du }
=> int {(u^2 -9)du} now i think it easy to int

don't forget to convert u to x after integral

Answer 2

Well Mac;

∫(x^3) / (√(x^2+9)) dx = (1/3) ( x^2 - 18) ( √(x^2 +9) +c

Good Luck..

Answer 3

Homework: let x=3tan(theta).

Looks like someone's calc II class has gotten to trig substitution.