# Integration by Parts Question: Integrate $x^3e^x$

Evaluate $$\int x^3e^x \mathrm{d}x$$ I tried to use integration by parts to do this and I let $u = x^3$ and $\mathrm{d}v = e^x \mathrm{d}x$. So I get $$\int x^3e^x \mathrm{d}x= \int x^3e^x \mathrm{d}x- \int e^x \cdot 3x^2\mathrm{d}x$$ How do I do this if my answer has the original integral in it?

• Your integration by parts formula is wrong. The first term on the right does not have an integral. – Wintermute Aug 14 '15 at 13:52

The formula for integration by parts is as follows: $$\int u \ \mathrm{d}v = uv - \int v \ \mathrm{d}u$$

For your integral $u = x^3$ and $\mathrm{d}v = e^x\mathrm{d}x$ so we get that $v = \int e^x \ \mathrm{d}x = e^x.$ $$\int x^3e^x \mathrm d x = x^3e^x - \int3x^2e^x \mathrm{d}x$$ Apply integration by parts a couple more times and you should get the correct answer.

A nice and quick way to visualize integration by parts (it could be a time-saver!):

$$\matrix{&\text{differentiate}&&& &\text{integrate}&\\ &x^3&&&&e^x&\\&&&\searrow^\color{red}{+}&&&\\&3x^2&&&&e^x&\\&&&\searrow^\color{blue}{-}&&&\\&6x&&&&e^x&\\&&&\searrow^\color{red}{+}&&&\\&6&&&&e^x&\\&&&\searrow^\color{blue}{-}&&&\\&0&&\rightarrow^{\color{red}{+}}&&e^x&}$$

You multiply the ends of each arrow, then place the corresponding sign and add everything up.

• This is not only *a nice and quick way * ! It is beautiful !! Thanks for providing it. – Claude Leibovici Mar 2 '16 at 9:25

Your next step isn't accurate. This should be the next:

$$\int x^3e^x \mathrm{d}x= x^3e^x \mathrm{d}x- \int e^x \cdot 3x^2\mathrm{d}x$$

Remember, integration by parts states: $\int uv' = uv - \int vu'$

It follows from the product rule:

$$(u \cdot v)' = uv' + vu'$$

Subtract $vu'$ from both sides:

$$(u \cdot v)' - vu' = uv'$$

Integrate both sides:

$$\int (u \cdot v)' - \int vu' = \int uv'$$

The integral and differentiation operators are, by definition, inverses of one another, so the formula becomes:

$$uv - \int vu' = \int uv'$$

Therefore, $\int uv' = uv - \int vu'$

$$x^n\int e^x\ dx\ne\int x^ne^x\ dx$$

$$I_n=\int x^ne^x\ dx=x^n\int e^x\ dx-\int\left[\dfrac{d(x^n)}{dx}\int e^x\ dx\right]dx$$

$$\implies I_n=x^ne^x-nI_{n-1}$$

Hint: Notice that you are applying it (Integration by parts Formula) wrong. It should go as

$$\int x^3 e^x dx = x^3 e^x - 3\int x^2 e^x dx$$

Apply it twice and see what you find.

make the ansatz $\int x^3e^xdx=e^x(Ax^3+Bx^2+Cx+D)$ with the real numbers $A,B,C,D$

If you're curious, there's a pattern that emerges by looking at $\int x^ne^x dx.$

It's a good induction exercise to prove the following:

$$\int x^ne^x dx = \left[ \sum^n_{j = 0}\left((-1)^j \frac{n!}{(n-j)!} x^{n-j}\right)\right]e^x.$$