Answer: \(\displaystyle \ln\left(\frac{2(a+1)^2}{a+2}\right)\).

Answer: \(\displaystyle \ln\left(\frac{2(a+1)^2}{a+2}\right)\).

Integrate term by term:

\(\int\left(\frac{2}{x+1}-\frac{1}{x+2}\right)\,dx =2\ln(x+1)-\ln(x+2).\)

Now apply the limits \(0\) and \(a\):

This gives

\(2\ln(a+1)-\ln(a+2)-\left(2\ln1-\ln2\right).\)

Since \(\ln1=0\), this becomes

\(2\ln(a+1)-\ln(a+2)+\ln2.\)

Use the laws of logarithms:

\(2\ln(a+1)=\ln\left((a+1)^2\right).\)

Therefore

\(2\ln(a+1)-\ln(a+2)+\ln2 =\ln\left(\frac{2(a+1)^2}{a+2}\right).\)