Recognise the structure before integrating

\int\frac{1}{2x}(\ln x)^5\,dx

Part	Integral
(a)	$\int\frac{1}{2x}(\ln x)^5\,dx$
(b)	$\int xe^{4-x^2}\,dx$
(c)	$\int(\ln x)^2\,dx$
(d)	$\int\frac4{1+\cos4\theta}\,d\theta$
(e)	$\int\frac1{\sin^2x}e^{\cot x}\,dx$
(f)	$\int\frac1{x^2+2x+5}\,dx$
(g)	$\int\frac{x^2-x}{x^2-16}\,dx$
(h)	$\int\frac{x^2-6x+1}{(3x+1)(x^2+3)}\,dx$
(i)	$\int\frac{e^{2x+1}}{(e^{2x}+1)^3}\,dx$
(j)	$\int\frac{x+1}{\sqrt{1-x^2}}\,dx$

2025 TJC H2 Maths (9758) JC2 MSM · original exercise attribution retained on source pages

Solution

(a)

Reverse chain rule

Evaluate $\int\frac1{2x}(\ln x)^5\,dx$ .

Explore this idea

Hint

Spot a function and its derivative: $\ln x$ and $\frac1x$ .
Use $\int f'(x)\,[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+C$ .

Answer · (a)

(\ln x)^6/12+C

Explore the idea

Does (ln x)⁶/12 track the area?

Move the two bounds to select an interval. Compare its signed integral with the change in the antiderivative.

Integration boundsa = 1.00, x = 2.50

Move either handle to choose the interval. The graphs compare the signed integral with F(x) − F(a).

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: 0.049

Below-axis contribution: 0

Signed integral: 0.049

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

Signed integral from 1.00 to x

0.049

F(x) − F(a)

0.049

agrees at displayed precision

Formula − integral

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

Differentiating $\frac{(\ln x)^6}{12}$ gives $\frac{(\ln x)^5}{2x}$ : the dashed curve and the solid curve agree.

Ideas to try

Move both bounds below 1: the integrand and signed integral are negative.
Set the lower bound to 1 and the upper bound to $e$ : what is the integral?

(b)

Reverse chain rule (exponential)

Evaluate $\int xe^{4-x^2}\,dx$ .

Explore this idea

Hint

The exponent's derivative is $-2x$ , which is a multiple of the other factor $x$ .
Use $\int f'(x)e^{f(x)}\,dx=e^{f(x)}+C$ .

Answer · (b)

-e^{4-x^2}/2+C

Explore the idea

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

0.00% 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

Slide to the left of 0: is the area negative or positive?
Why does the area stop growing for large $x$ ?

(c)

Integration by parts, twice

Evaluate $\int(\ln x)^2\,dx$ .

Explore this idea

Hint

There is no obvious reverse chain-rule pattern, so use parts with $dv=1\,dx$ .
The power of $\ln x$ drops after each step, so a second round finishes.

Answer · (c)

x(\ln x)^2-2x\ln x+2x+C

Explore the idea

Area under (ln x)²

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 = 2.50

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: 0.518

Below-axis contribution: 0

Signed integral: 0.518

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

Signed integral from 1.00 to x

0.518

antiderivative − F(1)

0.518

agrees at displayed precision

Formula − integral

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 three-term answer is the area function: the dashed curve lies on the solid one.

Ideas to try

x=1

the integrand is 0: why does the area flatten there?

(d)

Trigonometric identity

Evaluate $\int\dfrac4{1+\cos4\theta}\,d\theta$ .

Explore this idea

Hint

Use $\cos2A=2\cos^2A-1$ with $A=2\theta$ to remove the $1+$ .
The result is a $\sec^2$ integral.

Answer · (d)

\tan2\theta+C

Explore the idea

4/(1 + cos 4θ) and tan 2θ

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 = 0.35

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: 0.842

Below-axis contribution: 0

Signed integral: 0.842

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

Signed integral from 0.00 to x

0.842

tan 2θ − F(0)

0.842

agrees at displayed precision

Formula − integral

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

$\tan2\theta$ is the running area; both blow up at $\pm\frac\pi4$ , where $\cos4\theta=-1$ .

Ideas to try

Slide towards

\theta=\pm\frac\pi4

: what happens to the area?

(e)

Reverse chain rule (cot)

Evaluate $\int\dfrac1{\sin^2x}e^{\cot x}\,dx$ .

Explore this idea

Hint

Spot $\frac d{dx}\cot x=-\csc^2x=-\frac1{\sin^2x}$ in the integrand.

Answer · (e)

-e^{\cot x}+C

Explore the idea

e^(cot x)/sin²x on (0, π)

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 = 2.09

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: 0.432

Below-axis contribution: 0

Signed integral: 0.432

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

Signed integral from 1.57 to x

0.432

−e^(cot x) − F(π/2)

0.432

agrees at displayed precision

Formula − integral

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 $-e^{\cot x}$ is increasing because the integrand is positive.

Ideas to try

Slide towards

x=0

and towards

x=\pi

: which way does the area go?

(f)

Completing the square

Evaluate $\int\dfrac1{x^2+2x+5}\,dx$ .

Explore this idea

Hint

The quadratic has no real roots: complete the square.
Use $\int\frac{dx}{a^2+u^2}=\frac1a\tan^{-1}\frac ua+C$ .

Answer · (f)

\frac12\arctan((x+1)/2)+C

Explore the idea

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

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

0.00% 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 $\frac\pi2$ : the antiderivative levels off at $\pm\frac\pi4$ .

Ideas to try

The bell is centred at

x=-1

: why does the area change sign there?

(g)

Division then partial fractions

Evaluate $\int\dfrac{x^2-x}{x^2-16}\,dx$ .

Explore this idea

Hint

Numerator and denominator have the same degree: divide first.
Split the remainder into a log-derivative and a standard form.

Answer · (g)

x-\frac12\ln|x^2-16|+2\ln|\frac{x-4}{x+4}|+C

Explore the idea

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

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

0.00% 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 $-4<x<4$ the antiderivative matches; across a pole the constant of integration may change.

Ideas to try

The integrand has poles at

\pm4

. Stay between them and compare the curves.

(h)

Partial fractions

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

Explore this idea

Hint

The fraction is proper: write it as $\frac A{3x+1}+\frac{Bx+C}{x^2+3}$ .
Find $A,B,C$ by substitution.
Integrate each part.

Answer · (h)

\frac13\ln|3x+1|-\frac2{\sqrt3}\arctan(x/\sqrt3)+C

Explore the idea

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

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

0.00% 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 $A$ ) and an inverse tangent (from $-2$ ).

Ideas to try

Compare the sign of the area with the sign of the integrand.

(i)

Reverse chain rule with a power

Evaluate $\int\dfrac{e^{2x+1}}{(e^{2x}+1)^3}\,dx$ .

Explore this idea

Hint

Write $e^{2x+1}=e\cdot e^{2x}$ .
Spot $f=e^{2x}+1$ with $f'=2e^{2x}$ .

Answer · (i)

-e/[4(e^{2x}+1)^2]+C

Explore the idea

A peaked curve with a levelling 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 = 1.50

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: 0.168

Below-axis contribution: 0

Signed integral: 0.168

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

Signed integral from 0.00 to x

0.168

antiderivative − F(0)

0.168

agrees at displayed precision

Formula − integral

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 bounded: the integrand decays quickly on both sides.

Ideas to try

Why does the area barely change for large

x

(j)

Split, then standard forms

Evaluate $\int\dfrac{x+1}{\sqrt{1-x^2}}\,dx$ .

Explore this idea

Hint

Split into $\frac x{\sqrt{1-x^2}}$ and $\frac1{\sqrt{1-x^2}}$ .
The first is a reverse chain rule; the second is $\sin^{-1}x$ .

Answer · (j)

-\sqrt{1-x^2}+\arcsin x+C

Explore the idea

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

0.00% 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 $x=1$ is $\frac\pi2+1$ : an improper integral that converges.

Ideas to try

Slide towards

x=1

: the integrand blows up but the area stays finite.

Pull out the constant

Apply the power form

How the shaded areas add up

Match the derivative

Integrate

How the shaded areas add up

First round

Simplify

Second round

Combine

How the shaded areas add up

Rewrite the denominator

Simplify the integrand

Integrate

How the shaded areas add up

Match the derivative

Integrate

How the shaded areas add up

Complete the square

Composite arctangent rule

Integrate

How the shaded areas add up

Divide

Split the remainder

Integrate

Simplify

How the shaded areas add up

Set up

Find AAA

Find CCC and BBB

Integrate

How the shaded areas add up

Match f′f'f′

Integrate the power

How the shaded areas add up

Split

First part

Second part

Combine

How the shaded areas add up

Find $A$

Find $C$ and $B$

Match $f'$