Integrating a reciprocal linear function | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Recognise one over a linear expression

\frac{d}{dx}\ln|ax+b|=\frac{a}{ax+b}

\int\frac{1}{ax+b}\,dx

=\frac1a\ln|ax+b|+C

The shape must be a constant divided by a linearexpression.
The modulus bars allow the linear expression to be negative while keeping the logarithm's input positive.
If the denominator crosses zero inside a definite interval, the integral is not handled by simply substituting the limits.

Worked example

The coefficient outside and the coefficient inside

\int\frac{6}{3x-2}\,dx

The numerator 6 stays outside; dividing it by the inside coefficient 3 gives 2.

2\ln|3x-2|+C

Keep exact logarithms in definite integrals

Worked example

Current-paper area integral

For $y=27/(2x+5)$ from $x=0$ to $x=2$ :

\int_0^2\frac{27}{2x+5}\,dx

=\left[\frac{27}{2}\ln(2x+5)\right]_0^2

=\frac{27}{2}(\ln9-\ln5)

=\frac{27}{2}\ln\frac95

(9709/22/O/N/24 Q6)

Examiner note

If the question asks for an exact answer, leave logarithms in exact form. A calculator decimal loses the required form even when the method is correct.

Do not turn every fraction into a logarithm

Common mistake

\int1/(x^2+4)\,dx

is not

\ln(x^2+4)+C

. Differentiating that guess produces

2x/(x^2+4)

, not the original integrand.

Your turnIndependent check [4 marks]

Evaluate $\displaystyle\int_0^2\frac{9}{3x+1}\,dx$ exactly.

Show worked answer

\int\frac9{3x+1}\,dx

=3\ln(3x+1)

\left[3\ln(3x+1)\right]_0^2

=3\ln7-3\ln1

Answer

3\ln7

Recognise one over a linear expression

\frac{d}{dx}\ln|ax+b|=\frac{a}{ax+b}

\int\frac{1}{ax+b}\,dx

=\frac1a\ln|ax+b|+C

The shape must be a constant divided by a linearexpression.
The modulus bars allow the linear expression to be negative while keeping the logarithm's input positive.
If the denominator crosses zero inside a definite interval, the integral is not handled by simply substituting the limits.

Worked example

The coefficient outside and the coefficient inside

\int\frac{6}{3x-2}\,dx

The numerator 6 stays outside; dividing it by the inside coefficient 3 gives 2.

2\ln|3x-2|+C

Keep exact logarithms in definite integrals

Worked example

Current-paper area integral

For $y=27/(2x+5)$ from $x=0$ to $x=2$ :

\int_0^2\frac{27}{2x+5}\,dx

=\left[\frac{27}{2}\ln(2x+5)\right]_0^2

=\frac{27}{2}(\ln9-\ln5)

=\frac{27}{2}\ln\frac95

(9709/22/O/N/24 Q6)

Examiner note

If the question asks for an exact answer, leave logarithms in exact form. A calculator decimal loses the required form even when the method is correct.

Do not turn every fraction into a logarithm

Common mistake

\int1/(x^2+4)\,dx

is not

\ln(x^2+4)+C

. Differentiating that guess produces

2x/(x^2+4)

, not the original integrand.

Your turnIndependent check [4 marks]

Evaluate $\displaystyle\int_0^2\frac{9}{3x+1}\,dx$ exactly.

Show worked answer

\int\frac9{3x+1}\,dx

=3\ln(3x+1)

\left[3\ln(3x+1)\right]_0^2

=3\ln7-3\ln1

Answer

3\ln7