Q.Integrate the following function: $x \sqrt{1+2x^2}$

Concept: U Substitution — we look for an inner function whose derivative is present (up to a constant factor).

Let $u = 1 + 2x^2$. Then $du = 4x \, dx$, so $x \, dx = \frac{du}{4}$.

The integral becomes:

Integrate: …

Why U Substitution works here

When you see a function multiplied by its derivative (or a constant multiple of it), substitution is your best friend. Look at the integrand: $x \sqrt{1+2x^2}$. The inside of the square root is $1+2x^2$, and its derivative is $4x$. We have an $x$ sitting outside — not $4x$, but that’s fine; we can adjust for constants.

The square root is the "outer function," and $1+2x^2$ is the "inner function." Substitution lets us peel the layers: set $u$ equal to the inner function, replace everything in sight, and integrate a much simpler expression.

Step-by-step solution

1. Choose the substitution.

Let $u = 1 + 2x^2$. This is the expression inside the square root. The derivative is:

2. Solve for $x\,dx$.

We have $x\,dx$ in the original integral, not $4x\,dx$. So divide both sides by 4:

3. Rewrite the entire integral in terms of $u$.

The square root $\sqrt{1+2x^2}$ becomes $\sqrt{u}$. And $x\,dx$ becomes $\frac{1}{4}du$. So:

4. Integrate with respect to $u$. …

Method: Substitute when $x\,dx$ pairs with a quadratic under the root

Use this when a single $x$ multiplies a function of $x^2$ (or similar), e.g. $x\sqrt{1+2x^2}$: the $x\,dx$ is exactly what the substitution needs.

Steps

Step 1: Let $u$ be the expression under the root.

Put $u=1+2x^2$; then $du=4x\,dx$, so $x\,dx=\dfrac{du}{4}$.

Step 2: Rewrite the integral entirely in $u$.

$\int x\sqrt{1+2x^2}\,dx=\int u^{1/2}\cdot\frac{du}{4}=\frac14\int u^{1/2}\,du.$ …

Common Mistakes

Mistake 1: Forgetting the $\dfrac14$ from $du=4x\,dx$.

Why it's wrong: with $u=1+2x^2$, $x\,dx=\dfrac{du}{4}$; omitting the $\tfrac14$ scales the answer four times too large. Correct approach: solve $du=4x\,dx$ for $x\,dx$ before substituting.

Mistake 2: Choosing $u=x$ or $u=2x^2$ instead of the whole radicand. …