Video: Finding the Integration of a Function Involving an Exponential Function Using Integration by Parts Twice

Determine ∫2π‘₯²𝑒^(π‘₯ + 2) dπ‘₯.

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Video Transcript

Determine the indefinite integral of two π‘₯ squared times by 𝑒 to the π‘₯ plus two with respect to π‘₯.

First, we notice that we have a factor of two inside our integral. So we can factor this out of the integral. This leaves us with two times the indefinite integral of π‘₯ squared times by 𝑒 to the π‘₯ plus two with respect to π‘₯. In order to integrate this integral, we need to use integration by parts. When using integration by parts, we’re using the formula which tells us that the integral of 𝑒 times d𝑣 dπ‘₯ with respect to π‘₯ is equal to 𝑒 times 𝑣 minus the integral of d𝑒 dπ‘₯ times 𝑣 with respect to π‘₯.

From our integral, we need to choose a 𝑒 and a d𝑣 dπ‘₯. When picking our 𝑒 here, we want to pick a term which would decrease in magnitude when differentiated. And when picking our d𝑣 by dπ‘₯, we want to pick a term which would decrease or stay the same when integrated. And so therefore, we will pick 𝑒 equals π‘₯ squared. And d𝑣 by dπ‘₯ is equal to 𝑒 to the π‘₯ plus two.

So now, we can find d𝑒 by dπ‘₯ by differentiating 𝑒 with respect to π‘₯. And so that’s the same as differentiating π‘₯ squared with respect to π‘₯. So this gives us two π‘₯. And then, in order to find 𝑣, we simply integrate d𝑣 by dπ‘₯. And so that’s the same as integrating 𝑒 to the π‘₯ plus two with respect to π‘₯. And since this is an exponential term, this is simply 𝑒 to the π‘₯ plus two. So now, we have our integral in the form of the integral of 𝑒 times d𝑣 by dπ‘₯ dπ‘₯ since π‘₯ squared is 𝑒 and 𝑒 to the π‘₯ plus two is d𝑣 by dπ‘₯.

Now, we’re ready to use integration by parts. Using the formula and remembering our constant term, multiplying the integral, we get that this is equal to two times by π‘₯ squared 𝑒 to the π‘₯ plus two minus the integral of two π‘₯ 𝑒 to the π‘₯ plus two with respect to π‘₯. Now, we notice that we have another factor of two inside our integral, which can be moved and factored out of the integral. Next, we look at our integral again. And we notice that we can’t integrate this directly. So we will need to use integration by parts again.

So we can choose 𝑒 and d𝑣 by dπ‘₯ similarly to before, giving us 𝑒 equals π‘₯. And d𝑣 by dπ‘₯ is equal to π‘₯ plus two. Now, we can find d𝑒 by dπ‘₯ and 𝑣 by differentiating 𝑒 and integrating d𝑣 by dπ‘₯. This leaves us with d𝑒 by dπ‘₯ is equal to one. And 𝑣 is equal to 𝑒 to the π‘₯ plus two.

And now, we’re ready to use integration by parts again, remembering that our π‘₯ term is 𝑒 and our 𝑒 to the π‘₯ plus two term is d𝑣 by dπ‘₯. This leaves us with two times by π‘₯ squared 𝑒 to the π‘₯ plus two minus two times by π‘₯ 𝑒 to the π‘₯ plus two minus the integral of 𝑒 to the π‘₯ plus two with respect to π‘₯. And now, we have the integral of 𝑒 to the π‘₯ plus two with respect to π‘₯, which we know how to integrate.

Our next step here is to perform this integration. Since we’re integrating an exponential term here, it is just equal to itself. And so this leaves us with two times by π‘₯ squared 𝑒 to the π‘₯ plus two minus two π‘₯ times by 𝑒 to the π‘₯ plus two minus 𝑒 to the π‘₯ plus two. And now, since this was an indefinite integral, we mustn’t forget to add on our constant of integration 𝑐. Finally, we notice that we have an 𝑒 to the π‘₯ plus two, multiplying each term. And so we can factor this out here. Now, all that remains to do is to expand this minus two times by π‘₯ minus one bracket.

So we obtain a final answer that the indefinite integral of two π‘₯ squared times 𝑒 to the π‘₯ plus two with respect to π‘₯ is equal to two times by 𝑒 to the π‘₯ plus two times by π‘₯ squared minus two π‘₯ plus two plus 𝑐.

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