Question: Express $\dfrac{1}{x^2 - 1}$ in partial fractions and hence find $\displaystyle \int \dfrac{1}{x^2 - 1}\, dx$.

Step 1. Factor: $x^2 - 1 = (x - 1)(x + 1)$. Decompose:

$$\dfrac{1}{(x - 1)(x + 1)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$$

• $x = 1$: $\;1 = 2A \Rightarrow A = \dfrac{1}{2}$
• $x = -1$: $\;1 = -2B \Rightarrow B = -\dfrac{1}{2}$

So $\dfrac{1}{x^2 - 1} = \dfrac{1}{2(x - 1)} - \dfrac{1}{2(x + 1)}$.

Step 2. Integrate each term:

$$\int \dfrac{1}{x^2 - 1}\, dx = \dfrac{1}{2}\ln|x - 1| - \dfrac{1}{2}\ln|x + 1| + C$$

Combine using log laws:

$$= \dfrac{1}{2}\ln\!\left|\dfrac{x - 1}{x + 1}\right| + C$$

Tip: Each $\dfrac{1}{x + a}$ term integrates to $\ln|x + a|$. Log laws give a tidier final form.

Question: Express $\dfrac{1}{x(x + 1)^2}$ in partial fractions and find $\displaystyle \int \dfrac{1}{x(x + 1)^2}\, dx$.

Step 1. Form:

$$\dfrac{1}{x(x + 1)^2} \equiv \dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$$

Multiply through: $1 \equiv A(x + 1)^2 + Bx(x + 1) + Cx$.

• $x = 0$: $\;1 = A \Rightarrow A = 1$
• $x = -1$: $\;1 = -C \Rightarrow C = -1$
• Compare $x^2$: $\;0 = A + B \Rightarrow B = -1$

So $\dfrac{1}{x(x + 1)^2} = \dfrac{1}{x} - \dfrac{1}{x + 1} - \dfrac{1}{(x + 1)^2}$.

Step 2. Integrate:

$$\begin{aligned}\int \dfrac{1}{x(x + 1)^2}\, dx &= \ln|x| - \ln|x + 1| + \dfrac{1}{x + 1} + C \\ &= \ln\!\left|\dfrac{x}{x + 1}\right| + \dfrac{1}{x + 1} + C\end{aligned}$$

⚠ Watch out: Powers of $(x + a)$ integrate by the power rule to $-\dfrac{1}{(n - 1)(x + a)^{n - 1}}$ — not $\ln$. Only the first power gives $\ln$.

Question: Express $\dfrac{1}{r(r + 1)}$ in partial fractions, hence evaluate

$$\sum_{r = 1}^{n} \dfrac{1}{r(r + 1)}.$$

Step 1. Decompose:

$$\dfrac{1}{r(r + 1)} \equiv \dfrac{A}{r} + \dfrac{B}{r + 1}$$

Multiply: $1 \equiv A(r + 1) + Br$.

• $r = 0$: $\;1 = A$
• $r = -1$: $\;1 = -B \Rightarrow B = -1$

So $\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}$.

Step 2. Write out the sum:

$$\sum_{r = 1}^{n} \dfrac{1}{r(r + 1)} = \left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$$

Step 3. Telescoping — interior terms cancel:

$$\sum_{r = 1}^{n} \dfrac{1}{r(r + 1)} = 1 - \dfrac{1}{n + 1} = \dfrac{n}{n + 1}$$

Why this works: When consecutive partial-fraction terms differ by a shift (here $r$ and $r + 1$), summing makes a telescope — only the first and last terms survive.

Question: Express $f(x) = \dfrac{2x + 3}{(x - 1)(x + 2)}$ in partial fractions and hence find $f'(x)$.

Step 1. Decompose:

$$\dfrac{2x + 3}{(x - 1)(x + 2)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$$

• $x = 1$: $\;5 = 3A \Rightarrow A = \dfrac{5}{3}$
• $x = -2$: $\;-1 = -3B \Rightarrow B = \dfrac{1}{3}$

So $f(x) = \dfrac{5}{3(x - 1)} + \dfrac{1}{3(x + 2)}$.

Step 2. Differentiate each term ($\dfrac{d}{dx}(x + a)^{-1} = -(x + a)^{-2}$):

$$f'(x) = -\dfrac{5}{3(x - 1)^2} - \dfrac{1}{3(x + 2)^2}$$

Why decompose first: Differentiating $\dfrac{2x + 3}{(x - 1)(x + 2)}$ directly needs the quotient rule with a product denominator — messy. Decomposition turns it into two trivial power-rule derivatives.

Question: Express $\dfrac{3x + 1}{(x - 1)(x + 2)}$ in partial fractions.

Step 1. Form (each linear factor $\to A/(\text{factor})$):

$$\dfrac{3x + 1}{(x - 1)(x + 2)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$$

Step 2. Multiply both sides by $(x - 1)(x + 2)$:

$$3x + 1 \equiv A(x + 2) + B(x - 1)$$

Step 3. Cover-up — substitute strategic roots:

• $x = 1$: $\;3 + 1 = A(3) \Rightarrow A = \dfrac{4}{3}$
• $x = -2$: $\;-6 + 1 = B(-3) \Rightarrow B = \dfrac{5}{3}$

Answer:

$$\dfrac{3x + 1}{(x - 1)(x + 2)} = \dfrac{4}{3(x - 1)} + \dfrac{5}{3(x + 2)}$$

Tip: Cover-up only works for DISTINCT linear factors. Pick the $x$-value that zeros out the OTHER term.

Question: Express $\dfrac{2x^2 - 5x + 7}{(x - 1)(x - 2)(x + 1)}$ in partial fractions.

Step 1. Form:

$$\dfrac{2x^2 - 5x + 7}{(x - 1)(x - 2)(x + 1)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x - 2} + \dfrac{C}{x + 1}$$

Step 2. Multiply through:

$$2x^2 - 5x + 7 \equiv A(x - 2)(x + 1) + B(x - 1)(x + 1) + C(x - 1)(x - 2)$$

Step 3. Substitute each root:

• $x = 1$: $\;4 = A(-1)(2) \Rightarrow A = -2$
• $x = 2$: $\;5 = B(1)(3) \Rightarrow B = \dfrac{5}{3}$
• $x = -1$: $\;14 = C(-2)(-3) \Rightarrow C = \dfrac{7}{3}$

Answer:

$$\dfrac{-2}{x - 1} + \dfrac{5}{3(x - 2)} + \dfrac{7}{3(x + 1)}$$

Question: Express $\dfrac{x^3 + 2x^2}{(x - 1)(x + 2)}$ as a polynomial plus partial fractions.

Step 1. Check degrees: numerator $= 3$, denominator $= (x - 1)(x + 2) = x^2 + x - 2$ has degree $2$. Numerator $\geq$ denominator $\Rightarrow$ improper.

Step 2. Long division:

$$x^3 + 2x^2 \div (x^2 + x - 2) = (x + 1) + \dfrac{3x + 2}{x^2 + x - 2}$$

(Quotient $x + 1$, remainder $3x + 2$.)

Step 3. Now decompose the proper remainder:

$$\dfrac{3x + 2}{(x - 1)(x + 2)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$$

• $x = 1$: $\;5 = 3A \Rightarrow A = \dfrac{5}{3}$
• $x = -2$: $\;-4 = -3B \Rightarrow B = \dfrac{4}{3}$

Answer:

$$\dfrac{x^3 + 2x^2}{(x - 1)(x + 2)} = (x + 1) + \dfrac{5}{3(x - 1)} + \dfrac{4}{3(x + 2)}$$

⚠ Watch out: Partial-fraction setup fails (over-constrained) without dividing first when numerator degree $\geq$ denominator degree. Always check degrees up front.

Question: Express $\dfrac{2x^2 + 5x + 1}{(x + 1)(x + 2)}$ as a constant plus partial fractions.

Step 1. Denominator $= x^2 + 3x + 2$ (degree $2$). Numerator degree $2$. Equal degrees $\Rightarrow$ improper.

Step 2. Divide:

$$\dfrac{2x^2 + 5x + 1}{x^2 + 3x + 2} = 2 + \dfrac{(2x^2 + 5x + 1) - 2(x^2 + 3x + 2)}{x^2 + 3x + 2} = 2 + \dfrac{-x - 3}{(x + 1)(x + 2)}$$

Step 3. Decompose $\dfrac{-x - 3}{(x + 1)(x + 2)} \equiv \dfrac{A}{x + 1} + \dfrac{B}{x + 2}$.

Multiply through: $-x - 3 \equiv A(x + 2) + B(x + 1)$.

• $x = -1$: $\;-2 = A \Rightarrow A = -2$
• $x = -2$: $\;-1 = -B \Rightarrow B = 1$

Answer:

$$\dfrac{2x^2 + 5x + 1}{(x + 1)(x + 2)} = 2 + \dfrac{-2}{x + 1} + \dfrac{1}{x + 2}$$

Question: Express $\dfrac{8 - 3x}{(2x + 3)(x^2 + 4)}$ in partial fractions.

Step 1. Check $x^2 + 4$ is irreducible: $\Delta = 0 - 16 < 0$ ✓.

Form — irreducible quadratic factor needs a LINEAR numerator:

$$\dfrac{8 - 3x}{(2x + 3)(x^2 + 4)} \equiv \dfrac{A}{2x + 3} + \dfrac{Bx + C}{x^2 + 4}$$

Step 2. Multiply through:

$$8 - 3x \equiv A(x^2 + 4) + (Bx + C)(2x + 3)$$

Step 3. Find $A$ via cover-up at $x = -\tfrac{3}{2}$:

$$8 + \tfrac{9}{2} = A\!\left(\tfrac{9}{4} + 4\right) \;\Rightarrow\; \tfrac{25}{2} = \tfrac{25}{4} A \;\Rightarrow\; A = 2$$

Step 4. Compare $x^2$: $\;0 = A + 2B = 2 + 2B \Rightarrow B = -1$. Compare const: $\;8 = 4A + 3C = 8 + 3C \Rightarrow C = 0$.

Answer:

$$\dfrac{2}{2x + 3} + \dfrac{-x}{x^2 + 4}$$

⚠ Watch out: Confirm the quadratic is irreducible ($\Delta < 0$). If it factorises, treat the factors as two separate linear factors instead.

Question: Express $\dfrac{4x^2 + x + 1}{(x - 1)(x^2 + x + 1)}$ in partial fractions.

Step 1. Check $x^2 + x + 1$ is irreducible: $\Delta = 1 - 4 = -3 < 0$ ✓.

Form:

$$\dfrac{4x^2 + x + 1}{(x - 1)(x^2 + x + 1)} \equiv \dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + x + 1}$$

Step 2. Multiply through:

$$4x^2 + x + 1 \equiv A(x^2 + x + 1) + (Bx + C)(x - 1)$$

Step 3. Cover-up at $x = 1$:

$$4 + 1 + 1 = A(3) \;\Rightarrow\; A = 2$$

Step 4. Compare $x^2$: $\;4 = A + B = 2 + B \Rightarrow B = 2$. Compare const: $\;1 = A - C = 2 - C \Rightarrow C = 1$.

Answer:

$$\dfrac{2}{x - 1} + \dfrac{2x + 1}{x^2 + x + 1}$$

Question: Express $\dfrac{5x^2 + 3x + 7}{(x + 1)(x - 2)^2}$ in partial fractions.

Step 1. Set up — a repeated factor needs BOTH degrees:

$$\dfrac{5x^2 + 3x + 7}{(x + 1)(x - 2)^2} \equiv \dfrac{A}{x + 1} + \dfrac{B}{x - 2} + \dfrac{C}{(x - 2)^2}$$

Step 2. Multiply through:

$$5x^2 + 3x + 7 \equiv A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1)$$

Step 3. Use roots where possible, compare coefficients otherwise.

• $x = -1$: $\;5 - 3 + 7 = A(9) \Rightarrow A = 1$
• $x = 2$: $\;20 + 6 + 7 = C(3) \Rightarrow C = 11$
• Compare $x^2$: $\;5 = A + B = 1 + B \Rightarrow B = 4$

Answer:

$$\dfrac{1}{x + 1} + \dfrac{4}{x - 2} + \dfrac{11}{(x - 2)^2}$$

Why both degrees: $(x - 2)^2$ allows two independent constants — one for each power. Skipping $\dfrac{B}{x - 2}$ would lose a degree of freedom and over-constrain the system.

Question: Express $\dfrac{x^2 + 1}{(x - 1)^3}$ in partial fractions.

Step 1. A factor to power $n$ needs $n$ terms (powers $1, 2, \ldots, n$):

$$\dfrac{x^2 + 1}{(x - 1)^3} \equiv \dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2} + \dfrac{C}{(x - 1)^3}$$

Step 2. Multiply through by $(x - 1)^3$:

$$x^2 + 1 \equiv A(x - 1)^2 + B(x - 1) + C$$

Step 3. Expand and compare coefficients: RHS $= A x^2 + (-2A + B)x + (A - B + C)$.

• $x^2$: $\;A = 1$
• $x$: $\;-2A + B = 0 \Rightarrow B = 2$
• const: $\;A - B + C = 1 \Rightarrow C = 2$

Answer:

$$\dfrac{1}{x - 1} + \dfrac{2}{(x - 1)^2} + \dfrac{2}{(x - 1)^3}$$