Find the indefinite integral of x^2 times e^x with respect to x.

Step-by-step solution

1. 1

   Recognise the product type and pick $u$ so the polynomial shrinks. The integrand is a polynomial times an exponential, the classic signal for integration by parts:

   $\int u\,dv=uv-\int v\,du$

   Choose $u=x^2$ and $dv=e^x\,dx$. The reason is that differentiating $x^2$ lowers its degree, while $e^x$ is unchanged by integration — so each application of the formula moves us strictly closer to an integral we can do. Choosing the other way round ($u=e^x$) would raise the power to $x^3$ and make the problem worse.

2. 2

   First application of the formula. With $u=x^2$ and $dv=e^x\,dx$ we get $du=2x\,dx$ and $v=e^x$, so

   $\int x^2e^x\,dx=x^2e^x-\int e^x\cdot 2x\,dx=x^2e^x-2\int xe^x\,dx$

   The remaining integral has the same shape but with $x$ instead of $x^2$: the degree has dropped by one exactly as planned.

3. 3

   Second application, on $\int xe^x\,dx$. Take $u=x$, $dv=e^x\,dx$, so $du=dx$ and $v=e^x$:

   $\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x$

   In general $\int x^n e^x\,dx$ needs $n$ passes, because the degree falls by one each time. Here $n=2$, hence exactly two passes.

4. 4

   Substitute back and simplify. Feeding the second result into the first:

   $\int x^2e^x\,dx=x^2e^x-2\big(xe^x-e^x\big)=x^2e^x-2xe^x+2e^x$

   Factoring out the common $e^x$ gives the tidy form

   $\int x^2e^x\,dx=e^x\left(x^2-2x+2\right)+C$

   Do not forget the constant $C$: it is only added once, at the very end, not after each pass.

5. 5

   Verify by differentiating the answer. This is the one check that is always available for an indefinite integral. By the product rule,

   $\frac{d}{dx}\Big[e^x\big(x^2-2x+2\big)\Big]=e^x\big(x^2-2x+2\big)+e^x\big(2x-2\big)=e^x\big(x^2-2x+2+2x-2\big)=x^2e^x$

   The two linear pieces cancel and we recover the original integrand exactly, so the antiderivative is correct.

6. 6

   Note the pattern for reuse. The alternating-sign structure generalises:

   $\int x^ne^x\,dx=e^x\left(x^n-nx^{n-1}+n(n-1)x^{n-2}-\cdots\pm n!\right)+C$

   For $n=2$ this reads $e^x(x^2-2x+2)+C$, matching what we derived. Recognising this lets you write down $\int x^3e^x\,dx=e^x(x^3-3x^2+6x-6)+C$ without repeating the work three times.