Finding the integral

[kidmo87]

Hey everyone, I'm a little stuck on this problem. On my previous problems I've been finding the answers using inverse trig formula, but I feel like this problem will be different. Thanks.

 

[Deleted member 4993]

Hey everyone, I'm a little stuck on this problem. On my previous problems I've been finding the answers using inverse trig formula, but I feel like this problem will be different. Thanks.

You are correct - factor out common terms. Now integrate term by term [ i.e. integrate {x² dx} - integrate {1 dx}]

 

[kidmo87]

You are correct - factor out common terms. Now integrate term by term [ i.e. integrate {x² dx} - integrate {1 dx}]

I found the answer online when checking my work. I think I understand everything, except what I put in the red box.

 

[Deleted member 4993]

I think I understand everything, except what I put in the red box.

The integral should be \(\displaystyle \frac{1}{3}x(x^2-3) \ + \ C \)

 

[kidmo87]

Oh man, I'm sorry, that's what I meant. Forgot to put the square in it. Could you explain the part in the parenthesis please?

 

[HallsofIvy]

\(\displaystyle \frac{x^3}{3}- x= x\left(\frac{x^2}{3}- x\right)= x\left(\frac{x^2}{3}- \frac{3x}{3}\right)\)
now, also factor out the "\(\displaystyle \frac{1}{3}\)".

\(\displaystyle \frac{x}{3}\left(x^2- 3\right)\)