0606 P13 - Nov 2022 - Q11 - 6 marks
7882
It is given that
\(\int_1^a\left(\frac{3}{3x+2}-\frac{2}{2x+1}-\frac1x\right)\,dx=\ln\frac15,\)
where \(a\gt 1\). Find the exact value of \(a\).
Solution
Answer: \(\displaystyle a=2+\sqrt7\).
Answer: \(\displaystyle a=2+\sqrt7\).
First integrate:
\(\int\left(\frac{3}{3x+2}-\frac{2}{2x+1}-\frac1x\right)\,dx =\ln(3x+2)-\ln(2x+1)-\ln x.\)
Apply the limits \(1\) and \(a\):
\(\left[\ln(3x+2)-\ln(2x+1)-\ln x\right]_1^a=\ln\frac15.\)
At \(x=a\), this is
\(\ln(3a+2)-\ln(2a+1)-\ln a =\ln\left(\frac{3a+2}{a(2a+1)}\right).\)
At \(x=1\), this is
\(\ln5-\ln3-\ln1=\ln\frac53.\)
Therefore
\(\ln\left(\frac{3a+2}{a(2a+1)}\right)-\ln\frac53=\ln\frac15.\)
Combine the logarithms:
\(\ln\left(\frac{3(3a+2)}{5a(2a+1)}\right)=\ln\frac15.\)
So
\(\frac{3(3a+2)}{5a(2a+1)}=\frac15.\)
Multiplying by \(5a(2a+1)\),
\(3(3a+2)=a(2a+1).\)
Thus
\(9a+6=2a^2+a,\)
so
\(a^2-4a-3=0.\)
Solving,
\(a=2\pm\sqrt7.\)
Since \(a\gt 1\),
\(a=2+\sqrt7.\)
