00:01
Okay, so here we're going to be using integration by substitution, and we're going to let u equal x minus 1, so then du is equal to dx, and if u is equal to x minus 1, then x is going to be equal to u plus 1.
00:15
And so we're just going to plug that in for x and plug in u for x minus 1, and i'm going to put this up in the top right here.
00:23
And so this is equal to the integral of x is equal to u plus one divided by u to the one -half power.
00:32
And we also have d u.
00:34
And what we're going to do is we're just going to split this integral up into the integral of u divided by u to the one -half power, du plus the integral of one divided by u to the one -half power, du.
00:50
And so this first integral, u divided by the square root of u, that's just equal to the square root.
00:55
Of u we minus the power i'm in the denominator from that in the numerator and so we have the integral of the square root of u plus integral of the negative square of you or negative one -half power to u i guess negative square of you probably isn't the right way to say that it's u to the negative one -half power or one divided by the square root of you or the square of you yeah and so here now we can go ahead and just find the value of these two integrals or power rule integrals or antiderivatives so we add one to the power to the power.
01:26
So we're going to have you now to the three halves power and then we divide by that power or multiply by the reciprocal...