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S x²-x+1 x² + 2x + 2 2 -dx

S x²-x+1 x² + 2x + 2 2 -dx

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.2: Exponential Functions
Problem 3E
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Solve using the integration of rational function

08 - Rational Fractions.pdf 5 MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 1 How to resolve a proper rational fraction into its partial fractions. If a linear factor ax + b occurs once as a factor of the denominator, there corresponds to this factor a partial fraction of the from where A is a constant to be determined. A ax+b' • If a linear factor ax + b occurs n times as a factor of the denominator, there corresponds to this factor ʼn partial fractions in the form A₁ A₂ An + ax + b (ax + b)² (ax + b)n where A's are constants to be determined. MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 2 ● If an irreducible (not factorable) quadratic factor ax² + bx + c occurs once as a factor in the denominator, there corresponds to this factor one partial fraction in the form A(2ax+b)+B, where A and B are constants to be determined and 2ax + b is the derivative of ax² + bx + c. ax²+bx+c ¹ ● If an irreducible (not factorable) quadratic factor ax² + bx+c occurs n times as a factor in the denominator, there corresponds to this factor n partial fractions in the form A₁ (2ax + b) + B₁, ax² + bx + c + A₂(2ax + b) + B₂ (ax² + bx + c)² + An(2ax + b) + Bn (ax²+bx+c)n where A's and Bi's are constants to be determined and 2ax + b is the derivative of ax² + bx + c. 3 Sac Math138 - Engineering Calculus 2 3 How to solve the constants? ✪ By assigning a particular value to x so that some of the
Transcribed Image Text:08 - Rational Fractions.pdf 5 MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 1 How to resolve a proper rational fraction into its partial fractions. If a linear factor ax + b occurs once as a factor of the denominator, there corresponds to this factor a partial fraction of the from where A is a constant to be determined. A ax+b' • If a linear factor ax + b occurs n times as a factor of the denominator, there corresponds to this factor ʼn partial fractions in the form A₁ A₂ An + ax + b (ax + b)² (ax + b)n where A's are constants to be determined. MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 2 ● If an irreducible (not factorable) quadratic factor ax² + bx + c occurs once as a factor in the denominator, there corresponds to this factor one partial fraction in the form A(2ax+b)+B, where A and B are constants to be determined and 2ax + b is the derivative of ax² + bx + c. ax²+bx+c ¹ ● If an irreducible (not factorable) quadratic factor ax² + bx+c occurs n times as a factor in the denominator, there corresponds to this factor n partial fractions in the form A₁ (2ax + b) + B₁, ax² + bx + c + A₂(2ax + b) + B₂ (ax² + bx + c)² + An(2ax + b) + Bn (ax²+bx+c)n where A's and Bi's are constants to be determined and 2ax + b is the derivative of ax² + bx + c. 3 Sac Math138 - Engineering Calculus 2 3 How to solve the constants? ✪ By assigning a particular value to x so that some of the
S x²-x+1 x² + 2x + 2 -dx
Transcribed Image Text:S x²-x+1 x² + 2x + 2 -dx
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Can you solve using the same method shown on the first picture please

2x-10 dx = ( ( + = 1/5 ) dx X-3 = 12/1dx - 12/²3 dx = In • 3 ³ /x + ₁] - √n / x - 3] + C ] 2x-10 (x+1)(x-3) 2x-10 (X+1)(x-3) 2x-10 Let x= -1 -> A B t X + 1 X-3 A (x-3) + B(x+1) (x+1) (x-3) A (X-3) + B(x+1) 2(-1)-10 = A (-1-3) + B (-1+1) -12-4 A A = 3 H Let x = 3 2 (3)-10= A (3-3) + B (3-1) - 4 = 48 B = -.
Transcribed Image Text:2x-10 dx = ( ( + = 1/5 ) dx X-3 = 12/1dx - 12/²3 dx = In • 3 ³ /x + ₁] - √n / x - 3] + C ] 2x-10 (x+1)(x-3) 2x-10 (X+1)(x-3) 2x-10 Let x= -1 -> A B t X + 1 X-3 A (x-3) + B(x+1) (x+1) (x-3) A (X-3) + B(x+1) 2(-1)-10 = A (-1-3) + B (-1+1) -12-4 A A = 3 H Let x = 3 2 (3)-10= A (3-3) + B (3-1) - 4 = 48 B = -.
S x²-x+1 x² + 2x + 2 -dx
Transcribed Image Text:S x²-x+1 x² + 2x + 2 -dx
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