### Video Transcript

Find the integral of π₯ minus three timesed by π to the power of two π₯ with respect to π₯.

We can solve this integration by using integration by parts. Now, we know this because we have a polynomial which is π₯ minus three multiplied by an exponential which is π to the two π₯. And when performing integration by parts, we integrate one part of our integral and differentiate the other in order to form a new integral. And so, when we differentiate the polynomialβso thatβs π₯ minus threeβit will reduce. However, when we integrate the exponentialβthatβs π to the two π₯βit will only change by a constant and so it will not get any more difficult to integrate. Now, letβs quickly remind ourselves how integration by parts works.

If we have an integral of π’ multiplied by dπ£ by dπ₯ with respect to π₯, where π’ and dπ£ can be anything that we choose, then this integral is equal to π’π£ minus the integral of π£ timesed by dπ’ by dπ₯ with respect to π₯. So as we can see, the π’ from the original integral has been differentiated into the second integral to dπ’ by dπ₯. And the dπ£ by dπ₯ from the original integral has been integrated to π£ in the second integral. Now, letβs perform integration by part on our integral.

Weβll recall π₯ minus three π’. And π to the two π₯ will be dπ£ by dπ₯. Now, we need to find dπ’ by dπ₯ and π£. dπ’ by dπ₯ is the differential of π’ with respect to π₯. So we differentiate π₯ minus three with respect to π₯. In order to differentiate the π₯, we multiply by the power which is one since π₯ is equal to π₯ to the power of one and then reduce the power by one until we get π₯ to the zero. And π₯ to the zero is simply equal to one, giving us one multiplied by one. Then, since negative three is a constant and weβre differentiating it, this goes to zero. Since when differentiating any constant, we end up with zero. So therefore, dπ’ by dπ₯ is equal to one times one plus zero. So dπ’ by dπ₯ is simply equal to one.

Next, we need to find π£. And in order to do this, we simply integrate dπ£ by dπ₯, which is also equal to π to the power of two π₯. Weβll need to use the reverse chain rule in order to integrate this. We have that the integral of π to the power of π₯ with respect to π₯ is equal to π to the power of π₯. And so, when weβre integrating π to the power of two π₯, weβll get one divided by the differential of two π₯ which is two multiplied by π to the power of two π₯. This gives us one-half π to the power of two π₯.

Now remember that this two from the half has come from using the inverse chain rule. We have divided by the differential of two π₯. Now, we have found our π’, dπ’ by dπ₯, dπ£ by dπ₯, and π£. Weβre ready to use integration by parts. We have that the integral of π₯ minus three timesed by π to the two π₯ with respect to π₯ is equal to π’ timesed by π£. So thatβs π₯ minus three times one-half π to the two π₯ minus the integral of π£ dπ’ by dπ₯ with respect to π₯. So thatβs minus the integral of one-half π to the two π₯ times one with respect to π₯.

Now, this can be simplified to give π₯ minus three over two times π to the power of two π₯. And then, in the integral, we can ignore the times one since this has no effect in. And since one-half is a constant, we can bring it to the front of the integral giving us minus one-half times the integral of π to the two π₯ with respect to π₯.

Now, weβve already found the integral of π to the two π₯ with respect to π₯ since this is equal to π£. And we found that itβs equal to one-half π to the two π₯. This gives us π₯ minus three over two π to the two π₯ minus one-half times one-half π to the two π₯. And now, since this is the last integration we will perform, we mustnβt forget to add on our constant of integration or π, giving us plus π on the end. Next, we have one-half times one-half π to the two π₯. So we can write this as one-quarter π to the two π₯. And then, we notice that we have some π to two π₯ terms which we can group together.

In order to do this, we need to separate the fraction π₯ minus three over two into π₯ over two minus three over two. This gives us π₯ over two π the two π₯ minus three over two π to the two π₯ minus one-quarter π to the two π₯ plus π. When we simplify this, we find our solution which is that the integral of π₯ minus three π to the two π₯ with respect to π₯ is equal to π₯ over two times π to the two π₯ minus seven over four times π to the two π₯ plus π.