Answer: \(\displaystyle \frac12\ln\frac{27}{2}-\frac{1}{12}\).
Answer: \(\displaystyle \frac12\ln\frac{27}{2}-\frac{1}{12}\).
(a) Use the common denominator \((2x+1)^2(4x-1)\):
\(\frac{1}{2x+1} =\frac{(2x+1)(4x-1)}{(2x+1)^2(4x-1)},\)
\(-\frac{1}{(2x+1)^2} =-\frac{4x-1}{(2x+1)^2(4x-1)},\)
and
\(\frac{4}{4x-1} =\frac{4(2x+1)^2}{(2x+1)^2(4x-1)}.\)
Adding the numerators gives
\((2x+1)(4x-1)-(4x-1)+4(2x+1)^2.\)
Expand and simplify:
\(8x^2+2x-1-4x+1+16x^2+16x+4 =24x^2+14x+4.\)
Hence
\(\frac{1}{2x+1}-\frac{1}{(2x+1)^2}+\frac{4}{4x-1} =\frac{24x^2+14x+4}{(2x+1)^2(4x-1)}.\)
(b) Using part (a),
\(\int\frac{24x^2+14x+4}{(2x+1)^2(4x-1)}\,dx =\int\left(\frac{1}{2x+1}-\frac{1}{(2x+1)^2}+\frac{4}{4x-1}\right)dx.\)
Integrating term by term gives
\(\frac12\ln(2x+1)+\frac{1}{2(2x+1)}+\ln(4x-1).\)
Now apply the limits \(\frac12\) and \(1\):
\(\left[\frac12\ln(2x+1)+\frac{1}{2(2x+1)}+\ln(4x-1)\right]_{\frac12}^{1}.\)
At \(x=1\), this is
\(\frac12\ln3+\frac16+\ln3.\)
At \(x=\frac12\), this is
\(\frac12\ln2+\frac14+\ln1.\)
Since \(\ln1=0\), the integral is
\(\left(\frac12\ln3+\ln3+\frac16\right)-\left(\frac12\ln2+\frac14\right).\)
So
\(\frac32\ln3-\frac12\ln2-\frac{1}{12}.\)
Combine the logarithms:
\(\frac12\ln\left(\frac{27}{2}\right)-\frac{1}{12}.\)
