Q.Find $\int x\, e^x\, dx$

The key idea is integration by parts, which reverses the product rule. We choose $u$ and $dv$ so that $du$ is simpler and $v$ is easy to find.

Let $u = x$ and $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$.

Applying the formula $\int u\,dv = uv - \int v\,du$:

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

Why This Approach Works

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

Rearranging and integrating gives:

The trick is to pick $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $\int x e^x \, dx$, we want $u$ to be something that gets simpler when differentiated, and $dv$ to be something easy to integrate.

$x$ differentiates to $1$ (simpler), and $e^x$ integrates to itself (no harder). That’s the perfect match.

Step-by-Step Solution

1. Choose $u$ and $dv$

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

   Why? Differentiating $x$ gives $1$, which will simplify the next integral. Integrating $e^x$ gives $e^x$, which is just as easy to work with.

2. Find $du$ and $v$

   Differentiate $u$: $du = 1 \cdot dx = dx$.

   Integrate $dv$: $v = \int e^x \, dx = e^x$.

3. Apply the integration by parts formula

Substitute:

4. Evaluate the remaining integral

   $\int e^x \, dx = e^x + C$ (don’t forget the constant of integration).

5. Write the final result …

Method: Integration by Parts (Algebraic × Exponential)

Use this when a polynomial multiplies an exponential: differentiating the polynomial simplifies it while the exponential integrates unchanged.

Steps

Step 1: Apply the formula with LIATE.

Take $u=x$ (algebraic, simplifies on differentiating) and $dv=e^x\,dx$. Then $du=dx$, $v=e^x$. …

Common Mistakes

Mistake 1: Choosing $u=e^x$, $dv=x\,dx$.

Why it's wrong: differentiating $e^x$ never simplifies it, so the new integral is not easier. Correct approach: choose $u=x$ per LIATE.

Mistake 2: Sign error giving $xe^x+e^x$.

Why it's wrong: the formula subtracts $\int v\,du$, so the term is $-e^x$, giving $e^x(x-1)$. Correct approach: keep the minus sign. …