Recognise the structure before integrating

Evaluate:

PartIntegral
(a)∫12x(ln⁡x)5 dx\int\frac{1}{2x}(\ln x)^5\,dx
(b)∫xe4−x2 dx\int xe^{4-x^2}\,dx
(c)∫(ln⁡x)2 dx\int(\ln x)^2\,dx
(d)∫41+cos⁡4θ dθ\int\frac4{1+\cos4\theta}\,d\theta
(e)∫1sin⁡2xecot⁡x dx\int\frac1{\sin^2x}e^{\cot x}\,dx
(f)∫1x2+2x+5 dx\int\frac1{x^2+2x+5}\,dx
(g)∫x2−xx2−16 dx\int\frac{x^2-x}{x^2-16}\,dx
(h)∫x2−6x+1(3x+1)(x2+3) dx\int\frac{x^2-6x+1}{(3x+1)(x^2+3)}\,dx
(i)∫e2x+1(e2x+1)3 dx\int\frac{e^{2x+1}}{(e^{2x}+1)^3}\,dx
(j)∫x+11−x2 dx\int\frac{x+1}{\sqrt{1-x^2}}\,dx

2025 TJC H2 Maths (9758) JC2 MSM · original exercise attribution retained on source pages

Solution

(a)

Reverse chain rule

Evaluate ∫12x(ln⁡x)5 dx\int\frac1{2x}(\ln x)^5\,dx.

Explore this idea

Hint

  1. Spot a function and its derivative: ln⁡x\ln x and 1x\frac1x.

  2. Use ∫f′(x) [f(x)]n dx=[f(x)]n+1n+1+C\int f'(x)\,[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+C.

Answer · (a)

(ln⁡x)6/12+C(\ln x)^6/12+C.

x>0x>0.

Explore the idea

Open interactive exploration

Does (ln x)⁶/12 track the area?

Move the two bounds to select an interval. Compare its signed integral with the change in the antiderivative.

Integration boundsa = 1.00, x = 2.50

Move either handle to choose the interval. The graphs compare the signed integral with F(x) − F(a).

Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.049
Below-axis contribution
0
Signed integral
0.049

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 1.00 to x
0.049
F(x) − F(a)
0.049
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

Differentiating (ln⁡x)612\frac{(\ln x)^6}{12} gives (ln⁡x)52x\frac{(\ln x)^5}{2x}: the dashed curve and the solid curve agree.

Ideas to try
  1. Move both bounds below 1: the integrand and signed integral are negative.
  2. Set the lower bound to 1 and the upper bound to ee: what is the integral?