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

The key idea is U Substitution: the numerator $2x$ is exactly the derivative of the denominator $1+x^2$, so we set $u = 1+x^2$.

Step 1: Let $u = 1 + x^2$. Then $du = 2x \, dx$.

Step 2: The integral becomes

Step 3: Integrate: $\int \frac{1}{u} \, du = \log|u| + C$.

Step 4: Substitute back $u = 1 + x^2$. Since $1+x^2 > 0$ for all real $x$, the absolute value is unnecessary.

Why this works: the "derivative on top" pattern

When you see a fraction where the numerator is a constant multiple of the derivative of the denominator, you're looking at a logarithmic integral. The general rule is:

Here, the denominator is $1 + x^2$. Its derivative is $2x$, which is exactly the numerator. That's the green light — we can jump straight to the natural log.

Step-by-step solution

1. Spot the pattern. Let $u = 1 + x^2$. Then $du = 2x \, dx$. The numerator $2x \, dx$ is precisely $du$, so the integral becomes:

2. Integrate the simple form. The integral $\int \frac{du}{u}$ is one of the most basic results in calculus:

3. Substitute back. Replace $u$ with $1 + x^2$:

4. Simplify the absolute value (optional but tidy). Since $x^2 \ge 0$, we have $1 + x^2 \ge 1 > 0$ for all real $x$. The absolute value bars are unnecessary:

Method: Recognise the $\frac{f'(x)}{f(x)}$ log pattern (u-substitution)

Use this when the numerator is (a constant times) the derivative of the denominator, such as $\dfrac{2x}{1+x^2}$ where $2x=\dfrac{d}{dx}(1+x^2)$.

Steps

Step 1: Test whether the top is the derivative of the bottom.

Differentiate the denominator; if the numerator matches it (up to a constant factor), the integral is a logarithm.

Step 2: Substitute $u=$ denominator.

Let $u=1+x^2$, so $du=2x\,dx$. The integral becomes $\displaystyle\int\frac{du}{u}$.

Step 3: Integrate and back-substitute.

More generally $\displaystyle\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C$ — spotting this pattern saves the full substitution.

Common Mistakes

Mistake 1: Trying the power rule on $\dfrac{1}{1+x^2}$.

Why it's wrong: $1+x^2$ is not a single power of $x$, so $\dfrac{x^{n+1}}{n+1}$ does not apply. Correct approach: notice the numerator $2x$ is the derivative of $1+x^2$ and substitute.

Mistake 2: Missing the $\dfrac{f'}{f}$ structure.

Why it's wrong: overlooking that $2x=\dfrac{d}{dx}(1+x^2)$ leads to a stuck attempt. Correct approach: $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+C$, giving $\log(1+x^2)+C$.

Mistake 3: Writing $\dfrac{1}{2}\log(1+x^2)$ or similar extra factor.

Why it's wrong: here $du=2x\,dx$ matches the numerator exactly, so no $\tfrac12$ is needed. Correct approach: check that $du$ equals the numerator before inserting any constant.