Q.Integrate the following function: $\frac{2x}{(x^2+1)(x^2+3)}$

Concept: Partial Fraction Decomposition — splitting a rational function into simpler fractions whose denominators are the irreducible quadratic factors.

We want to integrate

Step 1: Decompose

Since both factors are irreducible quadratics, write

Step 2: Solve for constants

Multiply through by the denominator:

Comparing coefficients of $x^3$, $x^2$, $x$, and constant gives:

• $x^3$: $A + C = 0$
• $x^2$: $B + D = 0$
• $x$: $3A + C = 2$
• constant: $3B + D = 0$

From $A + C = 0$ and $3A + C = 2$, subtract to get $2A = 2 \Rightarrow A = 1$, then $C = -1$. …

Substitute. Let $u = x^2$, so $du = 2x\,dx$:

Partial fractions.

Integrate. …

Method: Spot the derivative-of-$x^2$ shortcut, then decompose in $u=x^2$

When a rational function contains only $x^2$ inside its factors and the numerator is a constant times $x$, the cleanest route is a substitution, not a full four-constant partial fraction.

Steps

Step 1: Check whether the numerator matches $\frac{d}{dx}(x^2)=2x$.

If the integrand is $\dfrac{(\text{const})\,x}{f(x^2)}$, set $u=x^2$ so that $du = 2x\,dx$. This absorbs the entire numerator and drops the problem one degree.

Step 2: Rewrite as a rational function in $u$.

Each factor $x^2+a$ becomes $u+a$, so the integral turns into $\displaystyle\int \frac{du}{(u+p)(u+q)}$ — distinct linear factors in $u$.

Step 3: Partial-fraction in $u$.

Use the identity for two distinct linear factors: …

Common Mistakes

Mistake 1: Setting up a four-constant decomposition when a substitution is far simpler.

Why it's wrong: With the numerator $2x$ exactly equal to $\frac{d}{dx}(x^2)$, the substitution $u=x^2$ collapses the whole problem — the $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3}$ setup wastes effort (and here yields $B=D=0$ anyway). Correct approach: Recognise $2x\,dx=du$ and reduce to $\int\frac{du}{(u+1)(u+3)}$.

Mistake 2: Using a constant numerator over an irreducible quadratic. …