Integration by Partial Fractions

Rational Functions: Proper, Improper, and the Plan

A rational function is a ratio $\frac{P(x)}{Q(x)}$ of polynomials with $Q(x) \neq 0$. It is proper if $\deg P < \deg Q$, otherwise improper.

The plan for integrating any rational function:

1. Improper? Divide first. Long division gives $\frac{P(x)}{Q(x)} = T(x) + \frac{P_1(x)}{Q(x)}$ with $T$ a polynomial (easy) and $\frac{P_1}{Q}$ proper.
2. Factorise the denominator into linear and quadratic factors.
3. Decompose the proper part into simpler fractions using Table 7.2.
4. Integrate each piece — they are all logs, powers, or $\tan^{-1}$ forms from Section 3.

Table 7.2 — the five decomposition forms:

Key Point: A repeated linear factor $(x-a)^2$ needs BOTH $\frac{A}{x-a}$ and $\frac{B}{(x-a)^2}$; an irreducible quadratic needs a full linear numerator $Bx + C$, not just a constant. These two rules are where wrong decompositions come from.

[Board Important] The decomposition is an identity (true for all $x$) — that is why both coefficient-comparison and value-plugging are valid ways to find the constants.

Finding the Constants — and Two Power Moves

Method 1 — compare coefficients. Clear denominators, expand, and match the coefficients of each power of $x$. Always works; more algebra.

Method 2 — plug convenient values. Since the identity holds for ALL $x$, substituting the roots kills terms instantly: in $1 = A(x+2) + B(x+1)$, putting $x = -1$ gives $A = 1$; putting $x = -2$ gives $B = -1$. Fast for distinct linear factors.

Power move 1 — substitution before decomposition. For trig rationals like $\int\frac{(3\sin\phi - 2)\cos\phi}{5 - \cos^2\phi - 4\sin\phi}\,d\phi$: substitute $y = \sin\phi$ first; the integral becomes an ordinary rational function $\int\frac{3y - 2}{(y - 2)^2}dy$ — then decompose.

Power move 2 — the $x^2 = y$ trick (decomposition only). For $\frac{x^2}{(x^2+1)(x^2+4)}$, treat $x^2$ as a single letter $y$ to decompose $\frac{y}{(y+1)(y+4)} = \frac{-1/3}{y+1} + \frac{4/3}{y+4}$ — then restore $x^2$ and integrate with $\tan^{-1}$ forms. The substitution is a bookkeeping device for the algebra; the integration still happens in $x$.

Key Point: Partial fractions is pure algebra followed by Section 1-3 integration. If your integration step feels hard, the decomposition step was wrong — go back.

[JEE Tip] After finding constants, spot-check with one easy value of $x$ (like $x = 0$) before integrating. Ten seconds of checking beats re-doing the whole problem.

Solved Examples

Example 1: Two distinct linear factors

Find $\displaystyle\int \frac{dx}{(x + 1)(x + 2)}$.

Solution:

1. Decompose: $\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$, so $1 = A(x+2) + B(x+1)$.
2. Find constants: $x = -1 \Rightarrow A = 1$; $x = -2 \Rightarrow B = -1$.
3. Integrate: $\int\frac{dx}{x+1} - \int\frac{dx}{x+2} = \log|x+1| - \log|x+2|$.

Final Answer: $\log\left|\frac{x + 1}{x + 2}\right| + C$.

Takeaway: Root-plugging found both constants in one line each — use it whenever the factors are distinct and linear.

Example 2: Improper — divide first

Find $\displaystyle\int \frac{x^2 + 1}{x^2 - 5x + 6}\,dx$.

Solution:

1. Degrees are equal (2 = 2) — improper. Divide: $\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \frac{5x - 5}{(x - 2)(x - 3)}$.
2. Decompose the proper part: $\frac{5x - 5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$; $5x - 5 = A(x - 3) + B(x - 2)$: $x = 2 \Rightarrow A = -5$; $x = 3 \Rightarrow B = 10$.
3. Integrate all three pieces: $\int dx - 5\int\frac{dx}{x-2} + 10\int\frac{dx}{x-3}$.

Final Answer: $x - 5\log|x - 2| + 10\log|x - 3| + C$.

Takeaway: Decomposing an improper fraction directly gives garbage — the division step is not optional.

Example 3: A repeated linear factor

Find $\displaystyle\int \frac{3x - 2}{(x + 1)^2(x + 3)}\,dx$.

Solution:

1. Decompose (Table form 4): $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$.
2. Clear and compare: $3x - 2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$. Comparing coefficients (or plugging $x = -1, -3$ plus one comparison): $A = \frac{11}{4}$, $B = -\frac{5}{2}$, $C = -\frac{11}{4}$.
3. Integrate: $\frac{11}{4}\log|x+1| - \frac{5}{2}\cdot\left(-\frac{1}{x+1}\right) - \frac{11}{4}\log|x+3|$.

Final Answer: $\frac{11}{4}\log\left|\frac{x+1}{x+3}\right| + \frac{5}{2(x+1)} + C$.

Takeaway: The $(x+1)^2$ piece integrates by the power rule to $-\frac{B}{x+1}$ — no logs from repeated-factor terms of power two.

Example 4: The $x^2 = y$ bookkeeping trick

Find $\displaystyle\int \frac{x^2}{(x^2 + 1)(x^2 + 4)}\,dx$.

Solution:

1. Decompose treating $x^2 = y$: $\frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$; $y = A(y+4) + B(y+1)$: $y = -1 \Rightarrow A = -\frac{1}{3}$; $y = -4 \Rightarrow B = \frac{4}{3}$.
2. Restore $x^2$: integrand $= -\frac{1}{3(x^2+1)} + \frac{4}{3(x^2+4)}$.
3. Integrate with $\tan^{-1}$ forms: $-\frac{1}{3}\tan^{-1}x + \frac{4}{3}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}$.

Final Answer: $-\frac{1}{3}\tan^{-1}x + \frac{2}{3}\tan^{-1}\frac{x}{2} + C$.

Takeaway: The $y$-substitution served only the algebra; the final integration used $x$ and the Section 3 forms. Never actually substitute $y = x^2$ in the integral itself (there is no $2x\,dx$ available!).

Example 5: Substitution first, then partial fractions

Find $\displaystyle\int \frac{(3\sin\phi - 2)\cos\phi}{5 - \cos^2\phi - 4\sin\phi}\,d\phi$.

Solution:

1. Substitute: $y = \sin\phi$, $dy = \cos\phi\,d\phi$; using $\cos^2\phi = 1 - y^2$, the denominator is $5 - (1 - y^2) - 4y = y^2 - 4y + 4 = (y - 2)^2$.
2. Decompose: $\frac{3y - 2}{(y-2)^2} = \frac{A}{y-2} + \frac{B}{(y-2)^2}$; $3y - 2 = A(y - 2) + B$: comparing, $A = 3$, $B = 4$.
3. Integrate: $3\log|y - 2| - \frac{4}{y - 2}$.
4. Return with $y = \sin\phi$ (and $2 - \sin\phi > 0$ always): $3\log(2 - \sin\phi) + \frac{4}{2 - \sin\phi} + C$.

Final Answer: $3\log(2 - \sin\phi) + \frac{4}{2 - \sin\phi} + C$.

Takeaway: Layered problems: substitution converts trig → rational, partial fractions finishes. Also note $|{\sin\phi - 2}| = 2 - \sin\phi$ since $\sin\phi \leq 1$.

Example 6: An irreducible quadratic factor

Find $\displaystyle\int \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)}\,dx$.

Solution:

1. Decompose (Table form 5): $\frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+1}$.
2. Compare: $x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2)$ gives $A + B = 1$, $2B + C = 1$, $A + 2C = 1$: $A = \frac{3}{5}$, $B = \frac{2}{5}$, $C = \frac{1}{5}$.
3. Split the quadratic piece for integration: $\frac{1}{5}\cdot\frac{2x + 1}{x^2+1} = \frac{1}{5}\cdot\frac{2x}{x^2+1} + \frac{1}{5}\cdot\frac{1}{x^2+1}$.
4. Integrate: $\frac{3}{5}\log|x+2| + \frac{1}{5}\log(x^2+1) + \frac{1}{5}\tan^{-1}x$.

Final Answer: $\frac{3}{5}\log|x + 2| + \frac{1}{5}\log(x^2 + 1) + \frac{1}{5}\tan^{-1}x + C$.

Takeaway: The $Bx + C$ numerator splits into a log part ($2x$ over the quadratic) and a $\tan^{-1}$ part (constant over the quadratic) — always in that order.

Example 7: The difference-of-squares special case

Find $\displaystyle\int \frac{dx}{x^2 - 9}$.

Solution:

1. Recognise the standard form (or decompose): $x^2 - 9 = (x-3)(x+3)$ with $a = 3$.
2. Apply: $\frac{1}{2 \times 3}\log\left|\frac{x - 3}{x + 3}\right| + C$.

Final Answer: $\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C$.

Takeaway: Section 3's formula (1) IS a pre-packaged partial fraction decomposition — recognising it saves the whole computation.

Example 8: Trig rational via $t = \sin x$

Find $\displaystyle\int \frac{\cos x}{(1 - \sin x)(2 - \sin x)}\,dx$.

Solution:

1. Substitute: $t = \sin x$, $dt = \cos x\,dx$: $\int\frac{dt}{(1 - t)(2 - t)}$.
2. Decompose: $\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}$; $1 = A(2 - t) + B(1 - t)$: $t = 1 \Rightarrow A = 1$; $t = 2 \Rightarrow B = -1$.
3. Integrate (watch signs): $\int\frac{dt}{1-t} = -\log|1-t|$ and $-\int\frac{dt}{2-t} = +\log|2-t|$.
4. Return: $\log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C$.

Final Answer: $\log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C$.

Takeaway: Factors of the form $(a - t)$ flip signs on integration: $\int\frac{dt}{a - t} = -\log|a - t|$. Track each minus separately.

Example 9: Improper with a cubic numerator

Find $\displaystyle\int \frac{x^3 + x + 1}{x^2 - 1}\,dx$.

Solution:

1. Divide: $\frac{x^3 + x + 1}{x^2 - 1} = x + \frac{2x + 1}{x^2 - 1}$.
2. Decompose: $\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$; $2x + 1 = A(x+1) + B(x-1)$: $x = 1 \Rightarrow A = \frac{3}{2}$; $x = -1 \Rightarrow B = \frac{1}{2}$.
3. Integrate: $\frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1|$.

Final Answer: $\frac{x^2}{2} + \frac{3}{2}\log|x - 1| + \frac{1}{2}\log|x + 1| + C$.

Example 10: An exponential rational

Find $\displaystyle\int \frac{dx}{e^x - 1}$.

Solution:

1. Substitute: $t = e^x$, $dt = e^x dx = t\,dx$, so $dx = \frac{dt}{t}$: the integral is $\int\frac{dt}{t(t - 1)}$.
2. Decompose: $\frac{1}{t(t-1)} = \frac{-1}{t} + \frac{1}{t-1}$ (plug $t = 0$ and $t = 1$).
3. Integrate: $-\log|t| + \log|t - 1| = \log\left|\frac{t-1}{t}\right|$.
4. Return: $\log\left|\frac{e^x - 1}{e^x}\right| = \log\left|1 - e^{-x}\right| + C$.

Final Answer: $\log\left|1 - e^{-x}\right| + C$.

Takeaway: $t = e^x$ turns exponential rationals into polynomial rationals — the extra $\frac{1}{t}$ from $dx = \frac{dt}{t}$ joins the denominator.