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

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

Clearing, $x = (Ax+B)(x-1) + C(x^2+1)$. Matching coefficients:

Integrate: …

Setting up the form

An irreducible quadratic factor needs a linear numerator; a linear factor needs a constant:

Step 1 — clear denominators and solve

Matching $x^2$, $x^1$, $x^0$ against $0\cdot x^2 + 1\cdot x + 0$:

From the last, $C=B$; from the first, $A=-C=-B$; then $B-A = 2B = 1 \Rightarrow B=\tfrac12$, so $A=-\tfrac12$, $C=\tfrac12$.

Step 2 — split the quadratic piece

Step 3 — integrate each term

• $\displaystyle -\frac12\int\frac{x}{x^2+1}\,dx = -\frac14\log(x^2+1)$ (let $u=x^2+1$). …

Method: Partial fractions with an irreducible quadratic factor

Use this for $\dfrac{N(x)}{(\text{irreducible quadratic})(\text{linear})}$, e.g. a factor $x^2+1$ that has no real roots. An irreducible quadratic factor needs a linear numerator, not a constant.

Steps

Step 1: Assign numerators by factor type. A linear factor $(x-r)$ gets a constant; an irreducible quadratic $(x^2+px+q)$ gets a linear numerator:

Step 2: Clear denominators and match all coefficients ($x^2$, $x^1$, $x^0$) to solve for $A,B,C$ (the cover-up trick only gives the constant $C$ directly). …

Common Mistakes

Mistake 1: Giving the irreducible quadratic a constant numerator.

Why it's wrong: writing $\frac{A}{x^2+1}$ instead of $\frac{Ax+B}{x^2+1}$ has too few unknowns and cannot reproduce the original fraction. Correct approach: an irreducible quadratic factor always takes a linear numerator $Ax+B$.

Mistake 2: Forgetting the $\tfrac12$ when integrating $\dfrac{x}{x^2+1}$.

Why it's wrong: $\int\frac{x}{x^2+1}dx=\frac12\log(x^2+1)$ (since $d(x^2+1)=2x\,dx$), whereas $\int\frac{1}{x^2+1}dx=\tan^{-1}x$ — mixing them scales the log term wrongly. Correct approach: split $\frac{Ax+B}{x^2+1}$ into the $x$-part (log) and the constant-part (arctan) separately. …