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Found in: Page 334

### Essential Calculus: Early Transcendentals

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

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