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 (f) Completing the square or open all working together.

(g)

Division then partial fractions

Evaluate ∫x2−xx2−16 dx\int\dfrac{x^2-x}{x^2-16}\,dx.

Explore this idea

Hint

  1. Numerator and denominator have the same degree: divide first.

  2. Split the remainder into a log-derivative and a standard form.

Answer · (g)

x−12ln⁡∣x2−16∣+2ln⁡∣x−4x+4∣+Cx-\frac12\ln|x^2-16|+2\ln|\frac{x-4}{x+4}|+C.

On intervals excluding ±4.

Explore the idea

Open interactive exploration

Between the poles at x = −4 and x = 4

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 = 1.90
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
0.011
Below-axis contribution
-0.049
Signed integral
-0.038

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

Signed integral from 0.00 to x
-0.038
antiderivative − F(0)
-0.038
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

Inside −4<x<4-4<x<4 the antiderivative matches; across a pole the constant of integration may change.

Ideas to try
The integrand has poles at ±4\pm4. Stay between them and compare the curves.