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 (e) Reverse chain rule (cot) or open all working together.

(f)

Completing the square

Evaluate ∫1x2+2x+5 dx\int\dfrac1{x^2+2x+5}\,dx.

Explore this idea

Hint

  1. The quadratic has no real roots: complete the square.

  2. Use ∫dxa2+u2=1atan⁡−1ua+C\int\frac{dx}{a^2+u^2}=\frac1a\tan^{-1}\frac ua+C.

Answer · (f)

12arctan⁡((x+1)/2)+C\frac12\arctan((x+1)/2)+C.

Explore the idea

Open interactive exploration

A bell curve and an inverse tangent

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 = 1.50
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
0.448
Below-axis contribution
0
Signed integral
0.448

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

Signed integral from -1.00 to x
0.448
½tan⁻¹((x+1)/2) − F(−1)
0.448
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 total area under the whole bell is π2\frac\pi2: the antiderivative levels off at ±π4\pm\frac\pi4.

Ideas to try
The bell is centred at x=−1x=-1: why does the area change sign there?