Question

Evaluate the integral:\(\int {\frac{y}{{(y + 4)(2y - 1)}}} dy\).

Short answer

The value of the integral is \(log{(y + 4)^{4/9}} + log{(2y - 1)^{1/18}} + c. \)

Step-by-step solution

Converting into partial fraction

\(\frac{y}{{(y + 4)\left( {2y - 1} \right)}} = \frac{A}{{(y + 4)}} + \frac{B}{{(2y - 1)}}\)

\(\frac{y}{{\left( {y + 4} \right)\left( {2y - 1} \right)}} = \frac{{A(2y - 1) + B(y + 4)}}{{\left( {y + 4} \right)\left( {2y - 1} \right)}}\)

\(y = A(2y - 1) + B(y + 4)\)

\(y = 2Ay - A + By + 4B\)

\(y = (2A + B)y + 4B - A\)

\(2A + B = 1\) ………(1)

And\(4B - A = 0\)

\(4B - A = 0 \Rightarrow A = 4B\)

Putting in equation (1)

\(2A + B = 1\)

\(2\left( {4B} \right) + B = 1\)

\(8B + B = 1\)

\(9B = 1 \Rightarrow B = \frac{1}{9}\)

And \(A = \frac{4}{9}\)

Finding The Integral

\(\int {\frac{{y\,dy}}{{(y + 4)(2y - 1)}}} = \int {\frac{{4/9}}{{y + 4}}} dy + \int {\frac{{1/9}}{{2y - 1}}} dy\)

Solving them separately.

\(\int {\frac{{4/9\,\,dy}}{{y + 4}}} = \frac{4}{9}\int {\frac{{dy}}{{y + 4}}} \)

Let

\(\begin{array}{l}y + 4 = u\\1 = \frac{{du}}{{dy}}\\ \Rightarrow dy = du\\\frac{4}{9}\int {1.\frac{{du}}{u} = \frac{4}{9}log\left| u \right|} \\ = \frac{4}{9}log\left| {y + 4} \right| + {c_1}\end{array}\)

Finalizing the Solution

\(\int {\frac{{\frac{1}{9}dy}}{{2y - 1}} = } \frac{1}{9}\int {\frac{{dy}}{{2y - 1}}} \)

Let

\(\begin{array}{l}2y - 1 = t\\2 = \frac{{dt}}{{dy}}\\dy = \frac{{dt}}{2}\\\frac{1}{9}\int {\frac{1}{t}.\frac{{dt}}{2} = \frac{1}{{18}}log\left| t \right|} \\ = \frac{1}{{18}}log\left| {2y - 1} \right| + {c_2}\end{array}\)

\(\frac{4}{9}log|y + 4| + {c_1} + \frac{1}{{18}}log|2y - 1| + {c_2}\)

Rearranging,

\(log{(y + 4)^{4/9}} + log{(2y - 1)^{1/18}} + c. \)\(\)