Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \(\int {\frac{{{x^2} + 1}}{{(x - 3){{(x - 2)}^2}}}} dx = 10\ln (1x - 31) - 9\ln (1x - 21) + \frac{5}{{x - 2}} + c\)

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Essential Calculus: Early Transcendentals

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Short Answer

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\)

Step by Step Solution

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

Step 2: Simplify and find the integral

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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\)

Step by Step Solution

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

Step 2: Simplify and find the integral

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