3.5 - Recognising a logarithmic integral | Cambridge A-Level Mathematics

Some fractions hide a familiar derivative: the numerator is a multiple of the denominator's derivative. In this part of 3.5, you will recognise that structure, adjust constants or rearrange a numerator, and integrate trigonometric examples such as tangent. Absolute values and interval checks will keep the logarithmic answers valid.

A derivative over its function

Why does $\int2x/(x^2+5)\,\mathrm dx$ have a logarithmic answer? The denominator is $f(x)=x^2+5$ , and its derivative is exactly the numerator $f'(x)=2x$ . The chain rule already tells us that $[\ln(x^2+5)]'=2x/(x^2+5)$ , so integration reverses that step.

You already know

For a differentiable positive function $f$ , the chain rule gives $[\ln f(x)]'=f'(x)/f(x)$ . For $f(x)<0$ , $\ln|f(x)|=\ln[-f(x)]$ , whose derivative is $[-f'(x)]/[-f(x)]=f'(x)/f(x)$ too.

Thus the same ratio works on either side of a zero, but never at the zero. Restrict each antiderivative to an interval where the denominator is nonzero; its arbitrary constant may differ between disconnected intervals.

A logarithmic integral

For differentiable $f$ with $f(x)\ne0$ throughout the interval, and a constant $k$ ,

\int k\frac{f'(x)}{f(x)}\,\mathrm dx=k\ln|f(x)|+C.

The case $k=1$ is in the formula booklet, with the arbitrary constant omitted there. A constant multiple follows by linearity of integration.

The function inside the logarithm is the denominator, not its derivative. If the denominator is always positive, the absolute-value bars may be omitted.

Worked example 1: a positive quadratic

Question. Find $\displaystyle\int\frac{2x}{x^2+5}\,\mathrm dx$ .

Plan. Identify the denominator and check that its derivative is the numerator.

Here $f=x^2+5$ and $f'=2x$ . Since $x^2+5>0$ for every real $x$ ,

\int\frac{2x}{x^2+5}\,\mathrm dx=\ln(x^2+5)+C.

Answer. $\ln(x^2+5)+C$ , for real $x$ .

Check. Differentiating gives $(x^2+5)'/(x^2+5)=2x/(x^2+5)$ , exactly the original integrand.

Change only the sign of the constant in the denominator. The same recognition works, but the denominator can now be negative or zero.

Worked example 2: a denominator that changes sign

Question. Find $\displaystyle\int\frac{2x}{x^2-5}\,\mathrm dx$ , stating the interval restriction.

Plan. Use the same derivative ratio, retain absolute values and exclude the denominator zeros.

The derivative of $x^2-5$ is $2x$ , hence

\int\frac{2x}{x^2-5}\,\mathrm dx=\ln|x^2-5|+C.

The denominator vanishes at $x=\pm\sqrt5$ .

Answer. $\ln|x^2-5|+C$ on any interval wholly within $x<-\sqrt5$ , $-\sqrt5<x<\sqrt5$ , or $x>\sqrt5$ .

Check. On the middle interval, the answer is $\ln(5-x^2)+C$ . Its derivative is $-2x/(5-x^2)=2x/(x^2-5)$ . On the outside intervals, it is $\ln(x^2-5)+C$ , with the same derivative ratio.

Taking an absolute value does not make the logarithm exist when the denominator is zero: $\ln0$ is still undefined. In particular, a finite ordinary definite integral must not be evaluated across such a singularity by merely subtracting two endpoint values.

Watch out

Do not write $\ln(x^2-5)$ on $-\sqrt5<x<\sqrt5$ . The original fraction is defined there, but that logarithm is not real. $\ln|x^2-5|$ supplies the valid real antiderivative.

Look for the derivative of a new quadratic denominator.

Finding the missing factor

The numerator need only be a constant multiple of the denominator's derivative. For $3x/(x^2+4)$ , the derivative of the denominator is $2x$ , so the missing constant is $3/2$ . You multiply the logarithm by that same constant.

Always compare the complete expressions, including coefficients. For example, $(6x+3)/(x^2+x+4)$ is three times the derivative ratio, because $6x+3=3(2x+1)$ . It therefore integrates to $3\ln(x^2+x+4)+C$ ; the denominator is positive since $x^2+x+4=(x+1/2)^2+15/4$ .

Worked example 3: a constant multiple

Question. Find $\displaystyle\int\frac{3x}{x^2+4}\,\mathrm dx$ .

Plan. Compare the numerator with $(x^2+4)'=2x$ , then factor out the ratio $3/2$ .

\begin{aligned} \int\frac{3x}{x^2+4}\,\mathrm dx &=\frac32\int\frac{2x}{x^2+4}\,\mathrm dx\\ &=\frac32\ln(x^2+4)+C \end{aligned}

Answer. $\tfrac32\ln(x^2+4)+C$ .

Check. Differentiating gives $(3/2)\times2x/(x^2+4)=3x/(x^2+4)$ , confirming the outside factor.

Now replace $3x$ by $3x^3$ , keeping the same denominator. The whole numerator is no longer a constant multiple of $2x$ . Rearrange it to leave a power term and a derivative ratio.

Worked example 4: exposing the ratio by rearrangement

Question. Find $\displaystyle\int\frac{3x^3}{x^2+4}\,\mathrm dx$ .

Plan. Write $x^2=(x^2+4)-4$ in the numerator so that one term cancels the denominator.

\begin{aligned} 3x^3&=3x(x^2+4)-12x\\ \frac{3x^3}{x^2+4}&=3x-\frac{12x}{x^2+4}\\ \int\frac{3x^3}{x^2+4}\,\mathrm dx &=\frac32x^2-6\ln(x^2+4)+C \end{aligned}

The first term uses the power rule; the second uses $-12x=-6(2x)$ .

Answer. $\tfrac32x^2-6\ln(x^2+4)+C$ .

Check. Differentiating and recombining gives

3x-\frac{12x}{x^2+4} =\frac{3x(x^2+4)-12x}{x^2+4} =\frac{3x^3}{x^2+4}.

This rearrangement is polynomial division expressed as an identity. It is useful when the numerator's degree is at least the denominator's. It does not make every fraction a logarithmic integral: for example, the numerator $1$ in $1/(x^2+4)$ is not a constant multiple of $2x$ .

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3.5 - Recognising a logarithmic integral

A derivative over its function

Finding the missing factor

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