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 (a) Reverse chain rule or open all working together.

(b)

Reverse chain rule (exponential)

Evaluate ∫xe4−x2 dx\int xe^{4-x^2}\,dx.

Explore this idea

Hint

  1. The exponent's derivative is −2x-2x, which is a multiple of the other factor xx.

  2. Use ∫f′(x)ef(x) dx=ef(x)+C\int f'(x)e^{f(x)}\,dx=e^{f(x)}+C.

Answer · (b)

−e4−x2/2+C-e^{4-x^2}/2+C.

Explore the idea

Open interactive exploration

An odd integrand and an even area

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 = 1.25
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
21.577
Below-axis contribution
0
Signed integral
21.577

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

Signed integral from 0.00 to x
21.577
−½e^(4−x²) − F(0)
21.577
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 antiderivative is a flipped bell shape; its slope matches the integrand.

Ideas to try
  1. Slide to the left of 0: is the area negative or positive?
  2. Why does the area stop growing for large xx?