Q.Integrate the function $x^2 e^x$

The key idea is integration by parts, applying the LIATE rule (Log-Inverse-Algebraic-Trig-Exp) to choose $u = x^2$ (algebraic) and $dv = e^x\,dx$ (exponential).

Step 1: Set $u = x^2$, $dv = e^x\,dx$. Then $du = 2x\,dx$, $v = e^x$.

Step 2: Apply integration by parts:

Step 3: For $\int 2x e^x\,dx$, again use parts: $u = 2x$, $dv = e^x\,dx$ → $du = 2\,dx$, $v = e^x$. This gives:

Step 4: Substitute back:

Why Integration by Parts?

When you see a product of two different kinds of functions — here, a polynomial ($x^2$) and an exponential ($e^x$) — the standard tool is integration by parts. It comes directly from the product rule for derivatives:

The trick is to choose $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For a polynomial times $e^x$, the polynomial gets simpler when differentiated, while $e^x$ stays the same when integrated. So we always set $u$ = polynomial, $dv = e^x dx$.

Step-by-Step Solution

1. First application of integration by parts

Let $u = x^2$ and $dv = e^x \, dx$. Then:

• $du = 2x \, dx$
• $v = e^x$

Applying the formula:

The new integral $\int x e^x \, dx$ is still a product of a polynomial ($x$) and $e^x$, but the polynomial degree has dropped from 2 to 1. We're making progress.

2. Second application of integration by parts

Now solve $\int x e^x \, dx$. Again, set $u = x$ and $dv = e^x \, dx$:

• $du = dx$
• $v = e^x$

So:

3. Substitute back

Plug the result from step 2 into step 1:

Simplify:

Since $-2C_1$ is just an arbitrary constant, we can rename it $C$:

Method: Integration by Parts, Applied Repeatedly (Polynomial × Exponential)

Use this when the integrand is a polynomial times an exponential (or similar), where one factor simplifies each time you differentiate it.

Steps

Step 1: Recall the formula and the LIATE choice.

Choose $u$ to be the factor that gets simpler on differentiating (the polynomial), and $dv$ the part easy to integrate (the exponential). LIATE (Log, Inverse-trig, Algebraic, Trig, Exponential) gives the priority for $u$.

Step 2: Differentiate down the polynomial.

Each application lowers the polynomial's degree by one. For $x^2 e^x$, take $u=x^2$, $dv=e^x dx$; the new integral $\int 2x e^x\,dx$ still has a polynomial, so apply parts again.

Step 3: Repeat until the polynomial becomes a constant.

Keep applying parts until the remaining integral is elementary, then collect terms:

Common Mistakes

Mistake 1: Choosing $u=e^x$ instead of $u=x^2$.

Why it's wrong: differentiating $e^x$ never simplifies it, so the integral does not reduce. Correct approach: pick $u$ as the algebraic factor (LIATE) so the polynomial degree drops.

Mistake 2: Sign errors when applying parts twice.

Why it's wrong: the formula has a minus sign, and a second application introduces another; mishandling them gives $x^2-2x-2$ instead of $x^2-2x+2$. Correct approach: bracket each stage and carry signs explicitly.

Mistake 3: Stopping after one application.

Why it's wrong: after the first step a polynomial factor still remains, so the integral is not yet elementary. Correct approach: repeat until only a constant multiplies $e^x$.