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

If this part uses an earlier result, revisit (b) Reverse chain rule (exponential) or open all working together.

(c)

Integration by parts, twice

Evaluate ∫(ln⁡x)2 dx\int(\ln x)^2\,dx.

Explore this idea

Hint

  1. There is no obvious reverse chain-rule pattern, so use parts with dv=1 dxdv=1\,dx.

  2. The power of ln⁡x\ln x drops after each step, so a second round finishes.

Answer · (c)

x(ln⁡x)2−2xln⁡x+2x+Cx(\ln x)^2-2x\ln x+2x+C.

x>0x>0.

Explore the idea

Open interactive exploration

Area under (ln x)²

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 2.50
Reset the controls

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
0.518
Below-axis contribution
0
Signed integral
0.518

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

Signed integral from 1.00 to x
0.518
antiderivative − F(1)
0.518
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

The three-term answer is the area function: the dashed curve lies on the solid one.

Ideas to try
At x=1x=1 the integrand is 0: why does the area flatten there?