00:01
In this problem, we have to find the indefinite integral of the expression 2 divided by e to the power negative x plus 1 and then followed by d x.
00:14
Now, in real life application, it is often useful to make everything in terms of the positive exponent.
00:23
And i have a negative exponent, e to the point negative x.
00:27
Let's make it a positive exponent as follows.
00:29
So this can be rewritten as 2 times e to the power x divided by e to the point negative x plus 1 multiplied by e to the power x dx.
00:42
So i'm multiplying both top and bottom simultaneously by e to the power x.
00:47
So then what do we achieve? we get 2 times e to the power x in the top and the bottom i get e to the power x plus 1.
00:58
And then dx so now i have to solve this problem which is exactly the same problem and the observation here is if i take the derivative of the bottom i get back my e to the power x and i see that there is the e to the power x hanging around the top of the expression so let's use that observation so let's take u is equal to e to the power x plus 1.
01:27
This implies d u is equal to e to the power x followed by d x.
01:33
So then this entire expression becomes two is outside.
01:39
Then e to the power x d x, let's go back...