Integral: Substitution prob: int (1 + x^3)^2 dx

[Chevy]

Here's the problem:

(1 + x^3)^2 dx

I've been using u so....

u = 1 + x^3 du = 3x^2 dx but dx = du/3x^2 ???

I got a bit confused around what dx should be. Maybe it's x^2 dx = du/3? But I wouldn't know how to use that.

I just know the answer should be x + (1/2)x^4 + (1/7)x^7 + C

Help would be appreciated.

 

[royhaas]

Expand the integrand and integrate termwise.

 

[N1CKY]

Here's the problem:

(1 + x^3)^2 dx

I've been using u so....

u = 1 + x^3 du = 3x^2 dx but dx = du/3x^2 ???

I got a bit confused around what dx should be. Maybe it's x^2 dx = du/3? But I wouldn't know how to use that.

I just know the answer should be x + (1/2)x^4 + (1/7)x^7 + C

Help would be appreciated.

(1 + x^3)^2 dx

1+x^3 = t => dx = 1/3(t-1)^2/3 dt


1/3

t^2(t-1)^(-2/3) dt

Let t-1 = p^3 => t=p^3+1 =? dt= 3p^2 dp

(p^3+1) dp

That is (p^4)/4 + p + Const

Now u should back trace it and get the answer

 

[galactus]

Just expand and integrate term by term. You're making it too hard.

INT((1+x^3)^2)dx=INT(1+2x^3+x^6)dx