00:01
Hello, so here we have the integral of x times the quantum, x squared minus 1, dx.
00:06
But if looking at x squared minus 1, the derivative there is just 2x.
00:13
And so if we do a u substitution and let u be equal to x squared minus 1 and then differentiate both sides, we get du is well equal to 2x dx, and then dividing both sides by 2, we get that 1ā2.
00:30
Over 2 or 1 1 1 half du is going to be equal to x dx.
00:35
So we can replace the x dx by 1 half du and then our integral is then going to be coming, we can pull out a 1 half in front and we get 1 half times the integral of while the x squared minus 1.
00:49
Is just going to be replaced by u.
00:50
So we just get the integral of 1 half times the integral of just u cubed du...