# Video: AP Calculus AB Exam 1 • Section I • Part A • Question 6

Calculate ∫𝑒^(𝑥) (1 − 𝑒^(−𝑥))² d𝑥.

04:03

### Video Transcript

Calculate the integral of 𝑒 to the 𝑥 power times one minus 𝑒 to the negative 𝑥 power squared d𝑥.

The first thing that we can do here is take this one minus 𝑒 to the negative 𝑥 power squared and expand it. We’re saying one minus 𝑒 to the negative 𝑥 power times itself. One minus 𝑒 to the negative 𝑥 power. Bring down the 𝑒 to the 𝑥 and the d𝑥. Inside this integral, we’re multiplying these three terms together. To simplify, we could multiply one minus 𝑒 to the negative 𝑥 times one minus 𝑒 to the negative 𝑥. And then, whatever we got there, we would multiply by 𝑒 to the 𝑥 power.

However, because we’re only dealing with multiplication, we can also multiply 𝑒 to the 𝑥 power times one minus 𝑒 to the negative 𝑥 power. And then from there, we’ll take whatever we get and multiply that by one minus 𝑒 to the negative 𝑥 power. Let’s do the second option.

We’ll distribute this 𝑒 to the 𝑥 power over one minus 𝑒 to the negative 𝑥, which is 𝑒 to the 𝑥, minus 𝑒 to the 𝑥 power times 𝑒 to the negative 𝑥 power. We have to be careful here. Remembering our power rules, if we’re multiplying 𝑎 to the 𝑚 power times 𝑎 to the 𝑛 power, what we do is add the exponents, 𝑚 plus 𝑛. In this case, that means we’ll be adding 𝑥 plus negative 𝑥. 𝑒 to the 𝑥 power times 𝑒 to the negative 𝑥 power equals 𝑒 to the zero power, which equals one.

We need to multiply 𝑒 to the 𝑥 power minus one times one minus 𝑒 to the negative 𝑥 power. 𝑒 to the 𝑥 power times one equals 𝑒 to the 𝑥. And now, we have 𝑒 to the 𝑥 power times negative 𝑒 to the negative 𝑥 power. So it looks like this: minus 𝑒 to the power times 𝑒 to the negative 𝑥 power, which we know is 𝑒 to the zero power. Then, we have negative one times one which is negative one and negative one times negative 𝑒 to the negative 𝑥 power which is positive 𝑒 to the negative 𝑥 power. Bringing down what we know, 𝑒 to the 𝑥 power minus 𝑒 to the zero — which is one — minus one plus 𝑒 to the negative 𝑥 power. One minus one equals negative two.

What we have here 𝑒 to the 𝑥 minus two plus 𝑒 to the negative 𝑥 is a simplified expression of what we started with. We can rewrite our original integral to say we wanna find the integral of 𝑒 to the 𝑥 power plus 𝑒 to the negative 𝑥 power minus two d𝑥. And this is much more manageable. We’ll just find the integral of each individual terms and add them together. The integral of 𝑒 to the 𝑥 d𝑥 equals 𝑒 to the 𝑥 power.

The integral of 𝑒 to the negative 𝑥 power with respect to 𝑥 is negative 𝑒 to the negative 𝑥 power. And we’re subtracting the integral of two with respect to 𝑥, which is negative two 𝑥 plus any constant 𝑐. The integral is 𝑒 to the 𝑥 power minus 𝑒 to the negative 𝑥 power minus two 𝑥 plus 𝑐. The key here was to simplify the expression that we were originally given into a format that was easier to integrate.