Q.Find $\int \dfrac{x^4\, dx}{(x-1)(x^2+1)}$

The numerator has higher degree than the denominator, so divide first, then use partial fractions on the proper remainder.

Step 1 — Divide. The denominator is $(x-1)(x^2+1)=x^3-x^2+x-1$. Dividing $x^4$ by it gives quotient $x+1$ and remainder $1$:

Step 2 — Decompose the remainder.

Put $x=1$: $1=2A\Rightarrow A=\tfrac12$. Compare $x^2$: $0=A+B\Rightarrow B=-\tfrac12$. Constant: $1=A-C\Rightarrow C=-\tfrac12$.

Step 3 — Integrate.

Why divide first

Partial fractions only apply to a proper rational function (numerator degree $<$ denominator degree). Here the numerator $x^4$ has degree $4$ while the denominator $(x-1)(x^2+1)=x^3-x^2+x-1$ has degree $3$, so we must divide before decomposing.

Step 1 — Polynomial long division

Divide $x^4$ by $x^3-x^2+x-1$:

• $x^4=x\,(x^3-x^2+x-1)+(x^3-x^2+x)$, giving a first quotient term $x$.
• $x^3-x^2+x=1\,(x^3-x^2+x-1)+1$, giving the next term $1$ and remainder $1$.

So the quotient is $x+1$ and the remainder is $1$:

Step 2 — Partial fractions on the remainder

The factor $x^2+1$ is irreducible, so it gets a linear numerator:

Clear denominators: $1=A(x^2+1)+(Bx+C)(x-1)$.

• Put $x=1$: $1=A(2)\Rightarrow A=\tfrac12$.
• Coefficient of $x^2$: $0=A+B\Rightarrow B=-\tfrac12$.
• Constant term: $1=A-C\Rightarrow C=A-1=-\tfrac12$.

(Check the $x$ coefficient: $-B+C=\tfrac12-\tfrac12=0$, as required.) Hence

Step 3 — Integrate term by term

Adding these gives the result.

Method: Improper rational function — divide first, then partial fractions

Use this whenever $\int\dfrac{P(x)}{Q(x)}\,dx$ has $\deg P \ge \deg Q$. Partial fractions only work on a proper fraction, so polynomial division is a mandatory first step.

Steps

Step 1: Check degrees; divide if top-heavy.

Here $\deg(x^4)=4$ exceeds $\deg\big((x-1)(x^2+1)\big)=3$, so do polynomial long division to write

where the remainder now has degree $<\deg Q$.

Step 2: Set up the partial-fraction form of the proper remainder.

Each distinct linear factor $(x-r)$ contributes $\dfrac{A}{x-r}$; each irreducible quadratic $(x^2+1)$ contributes a linear numerator $\dfrac{Bx+C}{x^2+1}$ (not just a constant).

Step 3: Solve for the constants.

Clear denominators and either substitute the real roots (fast for linear factors, e.g. $x=1$) or equate coefficients of like powers of $x$ to pin down $A,B,C$.

Step 4: Integrate term by term using standard forms.

$\int\tfrac{dx}{x-r}=\log|x-r|$, $\int\tfrac{x\,dx}{x^2+1}=\tfrac12\log(x^2+1)$, and $\int\tfrac{dx}{x^2+1}=\tan^{-1}x$. Add the polynomial's integral and a single $+C$.

Common Mistakes

Mistake 1: Applying partial fractions without dividing first.

Why it's wrong: $\dfrac{x^4}{(x-1)(x^2+1)}$ is improper ($\deg 4 \ge \deg 3$); decomposing it directly gives an inconsistent system. Correct approach: long-divide to get quotient $x+1$ and remainder $\dfrac{1}{(x-1)(x^2+1)}$, then decompose the remainder.

Mistake 2: Using a constant numerator over the quadratic factor.

Why it's wrong: an irreducible quadratic $x^2+1$ needs a linear numerator $Bx+C$, not just $\dfrac{B}{x^2+1}$; a constant loses a degree of freedom and the system won't solve. Correct approach: write $\dfrac{Bx+C}{x^2+1}$.

Mistake 3: Integrating $\dfrac{Bx+C}{x^2+1}$ as one log.

Why it's wrong: it must be split — $\dfrac{x}{x^2+1}$ gives $\tfrac12\log(x^2+1)$ while $\dfrac{1}{x^2+1}$ gives $\tan^{-1}x$; treating the whole thing as a log drops the arctangent term. Correct approach: separate the $x$-part (log) from the constant part (arctan).