Question

Evaluate the Integral.\(\int {\frac{{{x^2} - 5x + 16}}{{(2x + 1)\left( {x - 2} \right)}}} dx\)

Short answer

The solution of the given Integral is given as:

\(\frac{{3ln\left( {\left| {2x + 1} \right|} \right)}}{2} - ln\left( {\left| {x - 2} \right|} \right) - \frac{2}{{x - 2}} + C\)

Step-by-step solution

Use partial fraction decomposition.

\(\frac{{{x^2} - 5x + 16}}{{\left( {2x + 1} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{2x + 1}} + \frac{B}{{x - 2}} + \frac{C}{{{{(x - 2)}^2}}}\)

\({x^2} - 5x + 16 = A{\left( {x - 2} \right)^2} + B\left( {x - 2} \right)\left( {2x + 1} \right) + C(2x + 1)\)

\({x^2} - 5x + 16 = A\left( {{x^2} - 4x + 4} \right) + B\left( {2{x^2} - 3x - 2} \right) + C(2x + 1)\)

\({x^2} - 5x + 16 = A{x^2} - 4Ax + 4A + 2B{x^2} - 3Bx - 2B + 2Cx + C\)

\({x^2} - 5x + 16 = \left( {A + 2B} \right){x^2} + \left( { - 4A - 3B + 2C} \right)x + 4A - 2B\)

Compare the co-efficient and find A, B, C.

\(A + 2B = 1\)…… (1)

\( - 4A - 3B + 2C = - 5\) ….. (2)

\(C + 4A - 2B = 16\)…. (3)

From equation (1) & (3)

\(5A + C = 17\)….. (4)

\( \Rightarrow A = 1 - 2B\)….. (5)

Put value of equation (5) in equation (2) and (3)

\( - 4(1 - 2B) - 3B + 2C = - 5\) \(C + 4(1 - 2B) - 2B = 16\)

\( - 4 + 8B - 3B + 2C = - 5\) \(C + 4 - 8B - 2B = 16\)

\(5B + 2C = - 1\)………(6) \(C - 10B = 12\) ……(7)

From equation (7) and (6), we can write

\(5B + 2\left( {12 + 10B} \right) = - 1\)

\(5B + 24 + 20B = - 1\)

\(25B = - 25 \Rightarrow B = - 1\)

Substitute the value of B in equation (1) and (7)

\(A = 1 - 2\left( { - 1} \right) = 3\) \(C - 10\left( { - 1} \right) = 12\)

\(A = 3\) \(C = 2\)

Split up and find the integral.

\(\int {\frac{{{x^2} - 5x + 16}}{{\left( {2x + 1} \right){{\left( {x - 2} \right)}^2}}}} \,dx = \int {\frac{3}{{2x + 1}}} \,dx - \int {\frac{1}{{x - 2}}} \,dx + 2\int {\frac{1}{{{{(x - 2)}^2}}}\,} dx\)

\( = 3\frac{{ln(\mid \left| {2x + 1} \right|)}}{2} - ln(|x - 2|) - \frac{2}{{x - 2}} + C\)

Hence, the solution of the integral can be given as:

\(\int {\frac{{{x^2} - 5x + 16}}{{(2x + 1){{(x - 2)}^2}}}} dx = \frac{{3ln(\left| {2x + 1} \right|)}}{2} - ln(\left| {x - 2} \right|) - \frac{2}{{x - 2}} + C\)