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Q9E

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Essential Calculus: Early Transcendentals
Found in: Page 334
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Evaluate the integral: \(\int {\frac{{5x + 1}}{{(2x + 1)\left( {x - 1} \right)}}} dx\).

The value of the integral is \(log\left( {\sqrt {2x + 1} {{(x - 1)}^2}} \right) + c\)

See the step by step solution

Step by Step Solution

Step 1: Separating in the form of a partial fraction.

\(\frac{{5x + 1}}{{(2x + 1)(x - 1)}} = \frac{A}{{(2x + 1)}} + \frac{B}{{(x - 1)}}\)

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

Step 2:  Comparing both the sides:

\(5x + 1 = A(x - 1) + B(2x + 1)\)

\(5x + 1 = Ax - A + 2Bx + B\)

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

\( \Rightarrow A + 2B = 5\) ….(1)

And \( - A + B = 1\) ….(2)

Step 3: Solving equations 1 and 2.

\( - A + B = 1\)

\( \Rightarrow \;B = A + 1 \)

Putting in equation (1)

\(A + 2B = 5\)

\(A + 2(A + 1) = 5\)

\(A + 2A + 2 = 5\)

\(3A = 5 - 2\)

\(A = 1\)

\(B = A + 1 = 1 + 1\)

\(B = 2\)

Step 4: Finding the Equation

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

Solving the integrals separately.

Let \(2x + 1 = t\).

\(2 = \frac{{dt}}{{dx}} \Rightarrow dx = \frac{{dt}}{2}\)

So, \(\int {\frac{{1\,dx}}{{2x + 1}}} = \int {\frac{1}{t}} \frac{{dt}}{2} = \frac{1}{2}\int {\frac{1}{t}} dt\)

\( = \frac{1}{2}log|t| = \frac{1}{2}log|2x + 1| + {c_1}\) ……(1)

\(\int {\frac{2}{{x - 1}}} dx = 2\int {\frac{{dx}}{{x - 1}}} \)

Let \(x - 1 = v\)

\(l = \frac{{dv}}{{dx}} \Rightarrow dx = dv\)

\(2\int {\frac{{dv}}{v}} = 2log|v|\)

\( = 2log|x - 1| + {c_2}\) ……….(2)

So, \(\int {\frac{{5x + 1dx}}{{(2x + 1)(x - 1)}}} = \frac{1}{2}log|2x + 1| + {c_1} + 2log|x - 1| + {c_2}\).

(using equation (1) and (2))

Step 5: Rearranging

\(log{(2x + 1)^{1/2}} + log{(x - 1)^2} + {c_1} + {c_2}\)

\(\left( {as\,xlog\,x = log\,{x^n}} \right)\)

\( \Rightarrow log\left( {{{(2x + 1)}^{1/2}}{{(x - 1)}^2}} \right) + {c_1} + {c_2}\)

\(\left( {as\,\,loga + logb = logab} \right)\)

\( \Rightarrow log\left( {\sqrt {2x + 1} {{(x - 1)}^2}} \right) + c\)

\(\left( { as {c_1} + {c_2}} \right. is another constant i.e c)\)

Hence, The value of the integral is \(log\left( {\sqrt {2x + 1} {{(x - 1)}^2}} \right) + c\)

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