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 (g) Division then partial fractions or open all working together.

(h)

Partial fractions

Evaluate ∫x2−6x+1(3x+1)(x2+3) dx\int\dfrac{x^2-6x+1}{(3x+1)(x^2+3)}\,dx.

Explore this idea

Hint

  1. The fraction is proper: write it as A3x+1+Bx+Cx2+3\frac A{3x+1}+\frac{Bx+C}{x^2+3}.

  2. Find A,B,CA,B,C by substitution.

  3. Integrate each part.

Answer · (h)

13ln⁡∣3x+1∣−23arctan⁡(x/3)+C\frac13\ln|3x+1|-\frac2{\sqrt3}\arctan(x/\sqrt3)+C.

x≠−1/3x\ne-1/3.

Explore the idea

Open interactive exploration

Right of the pole at x = −1/3

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 = 3.50
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
Below-axis contribution
-0.327
Signed integral
-0.327

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

Signed integral from 1.00 to x
-0.327
antiderivative − F(1)
-0.327
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 answer combines a logarithm (from AA) and an inverse tangent (from −2-2).

Ideas to try
Compare the sign of the area with the sign of the integrand.