Q.Integrate the following function: $\frac{4x+1}{\sqrt{2x^2 + x - 3}}$

Concept: U Substitution — the numerator is almost the derivative of the expression inside the square root.

Let $u = 2x^2 + x - 3$. Then $du = (4x + 1)\,dx$, which matches the numerator exactly.

The integral becomes:

$\int \frac{4x+1}{\sqrt{2x^2 + x - 3}}\,dx = \int \frac{du}{\sqrt{u}} = \int u^{-1/2}\,du$ …

When you see an integral like this, the first instinct should be to check if the numerator is the derivative of the expression inside the square root. Why? Because the derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}} \cdot u'$, so if the numerator matches $u'$, the integral collapses into something simple.

Here, the denominator is $\sqrt{2x^2 + x - 3}$. Let’s call $u = 2x^2 + x - 3$. Then $du = (4x + 1)\,dx$. That’s exactly the numerator! So the integral becomes $\int \frac{du}{\sqrt{u}}$, which is a standard power rule.

Let’s walk through it step by step.

1. Set up the substitution.

   Let $u = 2x^2 + x - 3$. Then differentiate:

   $\frac{du}{dx} = 4x + 1$, so $du = (4x + 1)\,dx$.

2. Rewrite the integral.

   The original integral is $\int \frac{4x+1}{\sqrt{2x^2 + x - 3}}\,dx$.

   Substituting $u$ and $du$ gives:

   $\int \frac{du}{\sqrt{u}}$.

3. Integrate with respect to $u$.

   Recall that $\frac{1}{\sqrt{u}} = u^{-1/2}$. So:

   $\int u^{-1/2}\,du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$.

4. Substitute back.

   Replace $u$ with $2x^2 + x - 3$:

   $2\sqrt{2x^2 + x - 3} + C$. …

Method: When the numerator is the derivative of the radicand

If $\displaystyle\int \frac{f'(x)}{\sqrt{f(x)}}\,dx$, the whole integral collapses to $2\sqrt{f(x)}$ — no completing the square needed.

Steps

Step 1: Differentiate the expression under the root.

Compute $\dfrac{d}{dx}(ax^2+bx+c)=2ax+b$ and compare it with the numerator.

Step 2: Confirm the match (up to a constant factor). …

Common Mistakes

Mistake 1: Reaching for completing the square out of habit.

Why it's wrong: here the numerator is exactly the derivative of the radicand, so the problem is a one-line $u$-substitution — completing the square wastes time and invites errors. Correct approach: always check $\dfrac{d}{dx}(\text{radicand})$ against the numerator first.

Mistake 2: Forgetting the factor of $2$.

Why it's wrong: $\int u^{-1/2}\,du = 2u^{1/2}$, not $u^{1/2}$. Correct approach: apply the power rule $\int u^n\,du=\tfrac{u^{n+1}}{n+1}$ with $n=-\tfrac12$. …