P8.6Integration using partial fractions

• M1 for \(2x - 2 \equiv A(x + 5) + B(x + 2)\) and substituting a value of \(x\) or equating coefficients to find \(A\) or \(B\)
• A1 for one of \(A = -2\), \(B = 4\)
• A1 for \(-\frac{2}{x + 2} + \frac{4}{x + 5}\)

Worked solution: \(2x - 2 \equiv A(x + 5) + B(x + 2)\).
\(x = -2\): \(-6 = 3A\), so \(A = -2\). \(x = -5\): \(-12 = -3B\), so \(B = 4\).
So \(\frac{2x - 2}{(x + 2)(x + 5)} = -\frac{2}{x + 2} + \frac{4}{x + 5}\).

• M1 for \(\alpha\ln|x + 2| + \beta\ln|x + 5|\)
• A1ft for \(-2\ln|x + 2| + 4\ln|x + 5| + c\), following through their constants

Worked solution: \(\int \left(-\frac{2}{x + 2} + \frac{4}{x + 5}\right)\,\mathrm{d}x = -2\ln|x + 2| + 4\ln|x + 5| + c\)

• B1 for stating that \(\int \frac{1}{2x + 1}\,\mathrm{d}x = \frac{1}{2}\ln(2x + 1)\), not \(\ln(2x + 1)\) (the factor \(\frac{1}{2}\) from the chain rule is missing)

Worked solution: Differentiating \(\ln(2x + 1)\) gives \(\frac{2}{2x + 1}\), so \(\int \frac{1}{2x + 1}\,\mathrm{d}x = \frac{1}{2}\ln(2x + 1) + c\).

• M1 for \(3\ln(x + 1) - \frac{1}{2}\ln(2x + 1)\)
• dM1 for substituting the limits and using the laws of logarithms
• A1 for \(\ln\left(\frac{27\sqrt{3}}{8\sqrt{5}}\right)\) oe, e.g. \(\ln\left(\frac{27\sqrt{15}}{40}\right)\)

Worked solution: \(\left[3\ln(x + 1) - \frac{1}{2}\ln(2x + 1)\right]_{1}^{2} = \left(3\ln 3 - \frac{1}{2}\ln 5\right) - \left(3\ln 2 - \frac{1}{2}\ln 3\right) = \frac{7}{2}\ln 3 - \frac{1}{2}\ln 5 - 3\ln 2\)
\(= \ln\left(\frac{3^{\frac{7}{2}}}{5^{\frac{1}{2}} \times 8}\right) = \ln\left(\frac{27\sqrt{3}}{8\sqrt{5}}\right) = \ln\left(\frac{27\sqrt{15}}{40}\right)\)

• B1 for the correct form \(\frac{A}{x} + \frac{B}{2x + 1} + \frac{C}{(2x + 1)^{2}}\)
• M1 for \(4x^{2} + 6x + 3 \equiv A(2x + 1)^{2} + Bx(2x + 1) + Cx\) and a method to find at least one constant
• A1 for two correct constants
• A1 for \(\frac{3}{x} - \frac{4}{2x + 1} - \frac{2}{(2x + 1)^{2}}\)

Worked solution: \(\frac{4x^{2} + 6x + 3}{x(2x + 1)^{2}} = \frac{A}{x} + \frac{B}{2x + 1} + \frac{C}{(2x + 1)^{2}}\), so \(4x^{2} + 6x + 3 \equiv A(2x + 1)^{2} + Bx(2x + 1) + Cx\).
\(x = 0\): \(3 = A\), so \(A = 3\).
\(x = -\frac{1}{2}\): \(1 = -\frac{1}{2}C\), so \(C = -2\).
Comparing coefficients of \(x^{2}\): \(4 = 4A + 2B\), so \(B = -4\).

• M1 for integrating the first two terms to logarithms, with the factor \(\frac{1}{2}\) on \(\ln|2x + 1|\)
• M1 for \(\int \frac{C}{(2x + 1)^{2}}\,\mathrm{d}x = -\frac{C}{2(2x + 1)}\)
• A1ft for \(3\ln|x| - 2\ln|2x + 1| + \frac{1}{2x + 1}\)
• dM1 for substituting the limits and using the laws of logarithms
• A1 for \(6\ln 2 - 2\ln 3 - \frac{2}{9}\)

Worked solution: \(\int_{1}^{4} \left(\frac{3}{x} - \frac{4}{2x + 1} - \frac{2}{(2x + 1)^{2}}\right)\,\mathrm{d}x = \left[3\ln|x| - 2\ln|2x + 1| + \frac{1}{2x + 1}\right]_{1}^{4} = \left(\frac{1}{9} + 6\ln 2 - 4\ln 3\right) - \left(\frac{1}{3} - 2\ln 3\right) = 6\ln 2 - 2\ln 3 - \frac{2}{9}\)