Question: Show that $\dfrac{d}{dx}\left[(x-1)e^{-x}\right] = 2e^{-x} - x e^{-x}$. Hence find $\displaystyle\int x e^{-x}\,dx$.

Step 1. Differentiate $(x-1)e^{-x}$ by product rule:

$$\dfrac{d}{dx}\left[(x-1)e^{-x}\right] = 1 \cdot e^{-x} + (x-1)(-e^{-x}) = e^{-x} - (x-1)e^{-x} = e^{-x}(1 - x + 1) = e^{-x}(2 - x)$$

Or: $2e^{-x} - x e^{-x}$. $\checkmark$

Step 2. Integrate both sides (reverse):

$$\int (2e^{-x} - x e^{-x})\,dx = (x-1)e^{-x} + C$$ $$2\int e^{-x}\,dx - \int x e^{-x}\,dx = (x-1)e^{-x} + C$$

Step 3. Use $\int e^{-x}\,dx = -e^{-x}$:

$$-2e^{-x} - \int x e^{-x}\,dx = (x-1)e^{-x} + C$$ $$\int x e^{-x}\,dx = -(x-1)e^{-x} - 2e^{-x} + C = -(x+1)e^{-x} + C$$

Tip: AM's "Show ... hence find" is integration-by-parts in disguise. Always carry the constant of integration through.

Question: Express $\dfrac{5x^2 + 11x}{(4+x)(1+x)^2}$ in partial fractions. Hence integrate w.r.t. $x$.

Step 1. Form: $\dfrac{5x^2 + 11x}{(4+x)(1+x)^2} = \dfrac{A}{4+x} + \dfrac{B}{1+x} + \dfrac{C}{(1+x)^2}$.

Multiply through: $5x^2 + 11x = A(1+x)^2 + B(4+x)(1+x) + C(4+x)$.

Step 2. Cover-up:

• $x = -1$: $5 - 11 = -6 = C(3) \Rightarrow C = -2$.
• $x = -4$: $80 - 44 = 36 = A(9) \Rightarrow A = 4$.
• Coeff of $x^2$: $5 = A + B \Rightarrow B = 1$.

So $\dfrac{5x^2 + 11x}{(4+x)(1+x)^2} = \dfrac{4}{4+x} + \dfrac{1}{1+x} - \dfrac{2}{(1+x)^2}$.

Step 3. Integrate:

$$\int \dfrac{4}{4+x}\,dx + \int \dfrac{1}{1+x}\,dx - \int \dfrac{2}{(1+x)^2}\,dx$$ $$= 4\ln|4+x| + \ln|1+x| + \dfrac{2}{1+x} + C$$

Tip: For repeated linear factor $(1+x)^2$, the partial fractions split into TWO terms: $\dfrac{B}{1+x} + \dfrac{C}{(1+x)^2}$. The first gives $\ln$, the second gives a power.

Question: Express $\dfrac{3x}{3x+1}$ as $a + \dfrac{b}{3x+1}$ and hence find $\displaystyle\int \dfrac{3x}{3x+1}\,dx$.

Step 1. Rewrite by adding and subtracting $1$ in the numerator:

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

So $a = 1$, $b = -1$.

Step 2. Integrate term-by-term:

$$\int \dfrac{3x}{3x+1}\,dx = \int 1\,dx - \int \dfrac{1}{3x+1}\,dx = x - \dfrac{1}{3}\ln|3x+1| + C$$

Tip: When the integrand is a rational function where numerator degree $\geq$ denominator degree, long division (or this "add-and-subtract" trick) converts it to polynomial + simpler rational. Two flavours:

• $\dfrac{ax + b}{cx + d}$: split off a constant.
• Higher numerator: do polynomial long division.

Question: Show $(\cos x - \sin x)^2 = 1 - \sin 2x$. Hence find $\displaystyle\int_{\pi/2}^{\pi} (\cos x - \sin x)^2\,dx$.

Step 1. Expand:

$$(\cos x - \sin x)^2 = \cos^2 x - 2\sin x \cos x + \sin^2 x = (\cos^2 x + \sin^2 x) - 2\sin x \cos x = 1 - \sin 2x \;\checkmark$$

Step 2. Integrate:

$$\int (1 - \sin 2x)\,dx = x + \dfrac{1}{2}\cos 2x + C$$

Step 3. Evaluate from $\pi/2$ to $\pi$:

$$\left[x + \dfrac{1}{2}\cos 2x\right]_{\pi/2}^{\pi} = \left(\pi + \dfrac{1}{2}\cos 2\pi\right) - \left(\dfrac{\pi}{2} + \dfrac{1}{2}\cos \pi\right)$$ $$= \left(\pi + \dfrac{1}{2}\right) - \left(\dfrac{\pi}{2} - \dfrac{1}{2}\right) = \dfrac{\pi}{2} + 1$$

Tip: Standard identities for trig integration:

• $\sin^2 x = \dfrac{1 - \cos 2x}{2}$
• $\cos^2 x = \dfrac{1 + \cos 2x}{2}$
• $\sin 2x = 2\sin x \cos x$
• $\sec^2 x = 1 + \tan^2 x$

Always reduce to single-trig functions before integrating.

Question: $\dfrac{d^2y}{dx^2} = 5\cos 2x - 8\sin\dfrac{x}{2}$, the curve passes through $P(\pi, 7)$, gradient at $P$ is $3$. Find the equation of the curve.

Step 1. Integrate once:

$$\dfrac{dy}{dx} = \int (5\cos 2x - 8\sin\dfrac{x}{2})\,dx = \dfrac{5}{2}\sin 2x + 16\cos\dfrac{x}{2} + C_1$$

Step 2. Apply gradient condition at $x = \pi$:

$$3 = \dfrac{5}{2}\sin 2\pi + 16\cos\dfrac{\pi}{2} + C_1 = 0 + 0 + C_1 \Rightarrow C_1 = 3$$

So $\dfrac{dy}{dx} = \dfrac{5}{2}\sin 2x + 16\cos\dfrac{x}{2} + 3$.

Step 3. Integrate again:

$$y = -\dfrac{5}{4}\cos 2x + 32\sin\dfrac{x}{2} + 3x + C_2$$

Step 4. Apply point condition at $(\pi, 7)$:

$$7 = -\dfrac{5}{4}\cos 2\pi + 32\sin\dfrac{\pi}{2} + 3\pi + C_2 = -\dfrac{5}{4} + 32 + 3\pi + C_2$$ $$C_2 = 7 + \dfrac{5}{4} - 32 - 3\pi = -\dfrac{99}{4} - 3\pi$$

Answer: $y = -\dfrac{5}{4}\cos 2x + 32\sin\dfrac{x}{2} + 3x - \dfrac{99}{4} - 3\pi$.

Tip: Two integrations = two constants. Use the gradient condition to pin down $C_1$, then the point condition to pin down $C_2$.

Question: $\dfrac{dr}{dt} = -\dfrac{2}{(t+1)^2}$ with $r = 4$ when $t = 0$. Find $r$ in terms of $t$.

Step 1. Rewrite the integrand:

$$\dfrac{dr}{dt} = -2(t+1)^{-2}$$

Step 2. Integrate:

$$r = -2 \cdot \dfrac{(t+1)^{-1}}{-1} + C = \dfrac{2}{t+1} + C$$

Step 3. Apply initial condition $r(0) = 4$:

$$4 = \dfrac{2}{1} + C = 2 + C \Rightarrow C = 2$$

Answer: $r = \dfrac{2}{t+1} + 2$.

Tip: When the rate $\dfrac{dr}{dt}$ is given as a function of $t$ only, just integrate w.r.t. $t$ — no chain rule needed. The initial condition pins down the constant.

Question: Find $\displaystyle\int \left[\dfrac{5}{3x+1} - \dfrac{4}{(1-x)^2}\right]\,dx$.

Step 1. Apply standard formulas:

• $\int \dfrac{1}{ax+b}\,dx = \dfrac{1}{a}\ln|ax+b| + C$
• $\int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C$ (for $n \neq -1$)

Step 2. First term:

$$\int \dfrac{5}{3x+1}\,dx = \dfrac{5}{3}\ln|3x+1|$$

Step 3. Second term — rewrite $\dfrac{4}{(1-x)^2} = 4(1-x)^{-2}$:

$$\int 4(1-x)^{-2}\,dx = 4 \cdot \dfrac{(1-x)^{-1}}{(-1)(-1)} = \dfrac{4}{1-x}$$

(The two minus signs come from $n + 1 = -1$ and the inner derivative of $1 - x = -1$.)

Step 4. Combine:

$$\int \left[\dfrac{5}{3x+1} - \dfrac{4}{(1-x)^2}\right]\,dx = \dfrac{5}{3}\ln|3x+1| - \dfrac{4}{1-x} + C$$

Tip: Be careful with the inner-coefficient handling: for $\int (ax+b)^n$, divide by $a(n+1)$, NOT just $n+1$.