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

(j)

Split, then standard forms

Evaluate ∫x+11−x2 dx\int\dfrac{x+1}{\sqrt{1-x^2}}\,dx.

Explore this idea

Hint

  1. Split into x1−x2\frac x{\sqrt{1-x^2}} and 11−x2\frac1{\sqrt{1-x^2}}.

  2. The first is a reverse chain rule; the second is sin⁡−1x\sin^{-1}x.

Answer · (j)

−1−x2+arcsin⁡x+C-\sqrt{1-x^2}+\arcsin x+C.

−1<x<1-1<x<1.

Explore the idea

Open interactive exploration

Between x = −1 and x = 1

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 = 0.47
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
0.615
Below-axis contribution
0
Signed integral
0.615

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

Signed integral from 0.00 to x
0.615
antiderivative − F(0)
0.615
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 area to x=1x=1 is π2+1\frac\pi2+1: an improper integral that converges.

Ideas to try
Slide towards x=1x=1: the integrand blows up but the area stays finite.