arrrgggghhh!
Wolfram says:
http://integrals.wolfram.com/index.jsp?expr=x%5E2%2F%28%28x%2B3%29%283x%2B4%29%5E%281%2F2%29%29&random=false
integration of x^2/(x+3)sq.root of 3x+4 w.r.t. x
2 answers
Hmmm. Consider
u^2 = 3x+4, so x = (u^2-4)/3
2u du = 3 dx
x+3 = (u^2+5)/3
x^2 = (u^2-4)^2/9
x^2 / (x+3)√(3x+4) dx
= 2(u^2-4) / 9(u^2+5) du
= (2u^2-26)/9 + 18/(u^2+5)
Integrate that to get
2/27 u (u^2-39) + 18/√5 arctan(u/√5)
= 2/27 √(3x+4)(3x-35) + 18/√5 arctan(u/√5)
w00t!
u^2 = 3x+4, so x = (u^2-4)/3
2u du = 3 dx
x+3 = (u^2+5)/3
x^2 = (u^2-4)^2/9
x^2 / (x+3)√(3x+4) dx
= 2(u^2-4) / 9(u^2+5) du
= (2u^2-26)/9 + 18/(u^2+5)
Integrate that to get
2/27 u (u^2-39) + 18/√5 arctan(u/√5)
= 2/27 √(3x+4)(3x-35) + 18/√5 arctan(u/√5)
w00t!