00:01
Okay, choosing u equals x q plus x squared plus x plus 4, we know we can rewrite this integral as 1 over the square root of u, du, which is the same thing as u to the negative 1 1ā2, the integral of that, which is the same thing as 2 u to 1ā2 plus 0, which is equivalent to 2 square root of u.
00:25
Therefore, f prime x equals 2 x cubed plus x squared plus x plus 4, all under the radical.
00:41
And now graphing both these functions, the original one would look something like this, and then we're looking at the second one that looks like that, the one that we just found out 2 and then the radical.
00:59
Okay.
01:00
Now, in order to find the coefficient c, we have to evaluate the integral.
01:05
Therefore we can write the integral as from the bounds of three to x -cube plus x squared plus x plus four one over the square root of u d u therefore giving us two square root of u plus x squared plus x squared plus x plus four minus 2 and then the cube and then the square root of 3...