Q.Evaluate the definite integral: $\int_{2}^{3} \frac{x \, dx}{x^2+1}$

Concept: U Substitution — we substitute the denominator to simplify the integral into a standard logarithmic form.

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

Change the limits: when $x = 2$, $u = 2^2 + 1 = 5$; when $x = 3$, $u = 3^2 + 1 = 10$.

The integral becomes:

Evaluate: …

Why U-Substitution Works Here

When you see a function and its derivative lurking in the integrand, substitution is your best friend. In $\frac{x}{x^2+1}$, the numerator $x$ is (up to a constant factor) the derivative of the denominator $x^2+1$. That’s the classic signal: let $u$ be the denominator, and the integral collapses into a simple logarithmic form.

The key insight: you’re not just mechanically replacing variables — you’re undoing the chain rule. The derivative of $\log(x^2+1)$ is $\frac{2x}{x^2+1}$, so our integrand is half of that derivative. That’s the whole story.

Step-by-Step Solution

1. Choose the substitution.

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

2. Change the limits of integration.

   When $x = 2$, $u = 2^2 + 1 = 5$.

   When $x = 3$, $u = 3^2 + 1 = 10$.

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

4. Integrate. The antiderivative of $\frac{1}{u}$ is $\log|u|$. Since $u > 0$ on $[5,10]$, we drop the absolute value:

5. Simplify using logarithm properties. $\log 10 - \log 5 = \log\left(\frac{10}{5}\right) = \log 2$. …

Method: Substitution when the numerator is (a multiple of) the derivative of the denominator

Use this whenever a rational integrand $\dfrac{g'(x)}{g(x)}$ appears — the numerator matches the derivative of the denominator up to a constant, collapsing the integral to a logarithm.

Steps

Step 1: Spot the derivative pattern.

Differentiate the denominator mentally. If the numerator is a constant multiple of that derivative, the substitution $u = \text{denominator}$ will work. Here you would check $\dfrac{d}{dx}(x^2+1)=2x$ against the numerator $x$.

Step 2: Substitute and rescale the differential.

Set $u$ equal to the denominator, so $du = g'(x)\,dx$. Solve for the exact group that appears in the integrand (e.g. $x\,dx = \tfrac{1}{2}\,du$). The integral becomes a constant times $\displaystyle\int \frac{du}{u}$. …

Common Mistakes

Mistake 1: Keeping the original $x$-limits after substituting to $u$.

Why it's wrong: once the variable becomes $u=x^2+1$, the numbers $2$ and $3$ are $x$-values, not $u$-values. Correct approach: convert the limits to $u=5$ and $u=10$, or back-substitute $u=x^2+1$ before using the $x$-limits. …