Q.Integrate the following function: $\frac{x}{e^{x^2}}$

The key idea is U Substitution — we look for a function and its derivative inside the integrand.

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

The integral becomes:

Integrating: …

Why U Substitution Works Here

When you see an integral like $\int \frac{x}{e^{x^2}} \, dx$, the first thing to notice is the chain rule in reverse. The denominator $e^{x^2}$ is a composite function: the outer function is $e^{\text{(something)}}$, and the inner function is $x^2$. The derivative of that inner function, $2x$, is almost present in the numerator — we have $x$, not $2x$, but that’s a constant factor away. This is the classic signal for substitution: if you set $u$ equal to the “inside” part, the rest of the integrand (up to a constant) becomes $du$.

Let’s rewrite the integrand more cleanly:

Now it’s $x$ times $e^{-x^2}$. The derivative of $-x^2$ is $-2x$, so we’re off by a factor of $-2$. That’s fine — constants can be pulled out.

Step-by-Step Solution

1. Set up the substitution.

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

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

   The original integral is $\int x e^{-x^2} \, dx$. Substituting $x \, dx = \frac{du}{2}$ and $x^2 = u$, we get:

3. Integrate with respect to $u$. …

Method: Substitute the exponent when its derivative is an outside factor

Use this when an exponential of $x^2$ is multiplied by $x$, e.g. $\dfrac{x}{e^{x^2}}=x\,e^{-x^2}$: the $x$ supplies the derivative of the exponent.

Steps

Step 1: Rewrite as a clean exponential.

$\dfrac{x}{e^{x^2}}=x\,e^{-x^2}$.

Step 2: Substitute the exponent. …

Common Mistakes

Mistake 1: Leaving $\dfrac{x}{e^{x^2}}$ as a quotient.

Why it's wrong: it is easier to see the substitution once written as $x\,e^{-x^2}$. Correct approach: rewrite with a negative exponent first.

Mistake 2: Forgetting the $-\dfrac12$ from $du=-2x\,dx$.

Why it's wrong: with $u=-x^2$, $x\,dx=-\dfrac{du}{2}$; missing it changes both the sign and the size. Correct approach: substitute the differential exactly. …