Q.Integrate the following function: $\int x^2 e^{x^3} dx$ equals (A) $\frac{1}{3} e^{x^3} + C$ (B) $\frac{1}{3} e^{x^2} + C$ (C) $\frac{1}{2} e^{x^3} + C$ (D) $\frac{1}{2} e^{x^2} + C$

Concept: U Substitution — we look for a function whose derivative is also present in the integrand.

Notice that the derivative of $x^3$ is $3x^2$, and $x^2$ appears in the integrand.

Let $u = x^3$. Then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{1}{3} du$.

Substitute: …

The key to this problem is noticing that the derivative of $x^3$ is $3x^2$, and we have an $x^2$ sitting right next to the exponential. That’s the classic signal for a u-substitution: when you see a function and its derivative (up to a constant factor) multiplied together, substitution will cleanly undo the chain rule.

Let’s walk through it.

1. Choose the substitution.

   Let $u = x^3$. Why? Because the integrand contains $e^{x^3}$, and the derivative of $x^3$ is $3x^2$ — which is almost exactly the $x^2$ we have.

   Then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{1}{3} du$.

2. Rewrite the integral in terms of $u$.

   The original integral is $\int x^2 e^{x^3} dx$. Replace $x^3$ with $u$ and $x^2 dx$ with $\frac{1}{3} du$:

3. Integrate with respect to $u$. The integral of $e^u$ is simply $e^u + C$. So:

4. Substitute back to $x$. Since $u = x^3$, we get: …

Method: Direct $u$-substitution when the derivative is present

When one part of the integrand is (a constant multiple of) the derivative of another part inside a composite function, substitute that inner function.

Steps

Step 1: Spot the inner function and its derivative.

In $\int x^{2}e^{x^{3}}dx$, the exponent $x^{3}$ has derivative $3x^{2}$ — and an $x^{2}$ sits outside, so the substitution is set up for us.

Step 2: Let $t=x^{3}$, so $dt=3x^{2}\,dx$.

Hence $x^{2}\,dx=\dfrac{dt}{3}$.

Step 3: Rewrite entirely in $t$. …

Common Mistakes

Mistake 1: Substituting $t=x^{2}$ instead of $t=x^{3}$.

Why it's wrong: the exponent is $x^{3}$, so its derivative $3x^{2}$ is what cancels the outside $x^{2}$; using $x^{2}$ leaves an uncancelled $x$. This yields the distractor $\tfrac13 e^{x^{2}}$ (option B). Correct approach: substitute the exponent, $t=x^{3}$.

Mistake 2: Forgetting the $\dfrac13$ from $dt=3x^{2}\,dx$. …