Solve ∫x·eˣ dx using integration by parts LIATE rule.

Integration by Parts — ∫x·eˣ dx

Question

Evaluate the integral:

\int x \cdot e^x \, dx

This is a standard JEE Main question that tests whether you know when to apply integration by parts — and more importantly, how to pick the right function to differentiate.

Solution — Step by Step

When two functions are multiplied, integration by parts is the go-to method. The LIATE rule tells us which function to treat as $u$ (the one we differentiate):

Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential

Here we have $x$ (Algebraic) and $e^x$ (Exponential). Algebraic comes before Exponential in LIATE, so:

u = x, \quad dv = e^x \, dx

Differentiate $u$ and integrate $dv$ :

du = dx, \quad v = \int e^x \, dx = e^x

The reason we choose $x$ as $u$ is that differentiating it reduces the complexity — $x$ becomes $1$ , which kills the polynomial. If we did it the other way, we’d be integrating $x \cdot e^x$ again, which loops forever.

The formula is:

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

Substituting:

\int x \cdot e^x \, dx = x \cdot e^x - \int e^x \cdot dx

The remaining integral $\int e^x \, dx$ is straightforward:

= x \cdot e^x - e^x + C

Factor out $e^x$ for a cleaner final form.

Final Answer:

\boxed{\int x \cdot e^x \, dx = e^x(x - 1) + C}

Why This Works

Integration by parts is essentially the product rule for differentiation run in reverse. The product rule says $\frac{d}{dx}[uv] = u'v + uv'$ . Integrating both sides gives us $uv = \int u'v \, dx + \int uv' \, dx$ , which rearranges to the IBP formula.

The LIATE trick works because we want the $\int v \, du$ term to be simpler than what we started with. Differentiating a polynomial reduces its degree — differentiating $x$ gives $1$ , which collapses the remaining integral to just $\int e^x \, dx$ . This is why Algebraic beats Exponential in the priority order.

This pattern — polynomial × exponential — appears repeatedly in JEE Main and CBSE Class 12 boards. Once you’ve solved $\int x e^x dx$ , you can handle $\int x^2 e^x dx$ by applying IBP twice.

Alternative Method — Inspection / Reverse Product Rule

If you’ve done enough practice, you can spot the answer directly. Notice that:

\frac{d}{dx}\left[e^x(x-1)\right] = e^x(x-1) + e^x \cdot 1 = e^x \cdot x

So the integrand $x e^x$ is already the derivative of $e^x(x-1)$ . This “guess and verify” method saves time in multiple-choice questions where you just need the answer, not the working.

For any $\int x^n e^x dx$ , the answer always has the form $e^x \cdot (\text{polynomial of degree } n)$ . You can guess the form $e^x(ax + b)$ , differentiate, match coefficients, and get the answer in 20 seconds flat — no full IBP working needed in the exam hall.

Common Mistake

The most common error is swapping $u$ and $dv$ — taking $u = e^x$ and $dv = x \, dx$ . This gives $v = \frac{x^2}{2}$ , and now you’re left with $\int \frac{x^2}{2} e^x \, dx$ — which is harder than what you started with. You’ve gone from a degree-1 polynomial to degree-2. LIATE exists precisely to prevent this. When in doubt: differentiate the function that simplifies, integrate the one that stays manageable.

Question

Solution — Step by Step

Why This Works

Alternative Method — Inspection / Reverse Product Rule

Common Mistake

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