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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{{{x^2} + 1}}{{\leftEvaluate the integral: \(\int {\frac{{{x^2} + 1}}{{\left( {x - 3} \right){{\left( {x - 2} \right)}^2}}}} dx$$( {x - 3} \right){{\left( {x - 2} \right)}^2}}}} dx\)

The solution of the given integral is given as:

$$10\ln (1x - 31) - g\ln ( - x - 21) + \frac{5}{{x - 2}} + c$$

See the step by step solution

## Step 1: Write as Partial Fraction and find A, B, C

$$\frac{{{x^2} + 1}}{{\left( {x - 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{x - 3}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}}$$

$$\therefore {x^2} + 1 = \left( {\left( {x - 3} \right){{\left( {x - 2} \right)}^2}} \right)\left( {\frac{A}{{x - 3}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 3} \right)}^2}}}} \right)$$

$$\therefore {x^2} + 1 = A{(x - 2)^2} + B(x - 2)(x - 3) + C(x - 3)$$

$$\begin{array}{l}\therefore {x^2} + 1 = A({x^2} - 4x + 4) + B({x^2} - 5x + 6) + C(x - 3)\\\therefore {x^2} + 1 = A{x^2} - 4Ax + 4A + B{x^2} - 5Bx + 6B + Cx - 3C\\\therefore {x^2} + 1 = (A{x^2} + B{x^2}) + ( - 4Ax - 5Bx + Cx) + 4A + 6B - 3C\\\therefore {x^2} + 1 = (A + B){x^2} + ( - 4A - 5B + C)x + 4A + 6B - 3C\end{array}$$

By comparing the coefficients of $${x^2}$$and $$x$$

$$\begin{array}{l}\therefore A + B = 1 \to (1)\\4A + 6B - 3C = 1 \to (2)\\ - 4A - 5B + C = 0 \to (3)\end{array}$$

Solve the system of equations,

$$\therefore A + B = 1 \Rightarrow A = 1 - B$$

Put the value of A in equation (3)

$$\begin{array}{l}\therefore - 4(1 - B) - 5B + C = 0 \Rightarrow - 4 + 4B - 5B + C = 0\\\therefore - 4 - B + C = 0 \Rightarrow C - B = 4 \to (4)\end{array}$$

Now, put $$C = 4 + B$$ in equation (2)

$$\begin{array}{l}\therefore 4A + 6B - 3(4 + B) = 1 \Rightarrow 4A + 6B - 12 - 3B = 1\\\therefore 4A + 3B = 13 \to (5)\end{array}$$

Put $$A = 1 - B$$ in equation (5),

We get $$4(1 - b) + 3b = 13$$

Substitute value of B in equations (4) and (1)

## Step 2: Simplify and find the integral

$$\therefore \frac{{{x^2} + 1}}{{\left( {x - 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{{10}}{{x - 3}} + \frac{{ - 9}}{{x - 2}} + \frac{{ - 5}}{{{{(x - 2)}^2}}}$$

Apply integral both sides

$$\int {\frac{{{x^2} + 1}}{{\left( {x - 3} \right){{\left( {x - 2} \right)}^2}}}} = \int {\frac{{10}}{{x - 3}}} - 9\int {\frac{1}{{x - 2}}dx} - 5\int {\frac{1}{{{{\left( {x - 2} \right)}^2}}}dx}$$

$$\therefore \int {\frac{{{x^2} + 1}}{{\left( {x - 3} \right){{\left( {x - 2} \right)}^2}}}} dx = 10\ln (1x - 31) - 9\ln (1x - 2) - 5\left( { - \frac{1}{{x - 2}}} \right) + c$$

$$\therefore \int {\frac{{{x^2} + 1}}{{(x - 3){{(x - 2)}^2}}}} dx = 10\ln (1x - 31) - 9\ln (1x - 2) + \frac{5}{{x - 2}} + c$$

Hence, the solution of the integral can be given as $$\int {\frac{{{x^2} + 1}}{{(x - 3){{(x - 2)}^2}}}} dx = 10\ln (1x - 31) - 9\ln (1x - 21) + \frac{5}{{x - 2}} + c$$