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Ex 7.5, 13 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Integration by partial fraction - Type 5
Chapter 7 Class 12 Integrals
Concept wise
- Using Formulaes
- Using Trignometric Formulaes
- Integration by substitution - x^n
- Integration by substitution - lnx
- Integration by substitution - e^x
- Integration by substitution - Trignometric - Normal
- Integration by substitution - Trignometric - Inverse
- Integration using trigo identities - sin^2,cos^2 etc formulae
- Integration using trigo identities - a-b formulae
- Integration using trigo identities - 2x formulae
- Integration using trigo identities - 3x formulae
- Integration using trigo identities - CD and CD inv formulae
- Integration using trigo identities - Inv Trigo formulae
- Integration by parts
- Integration by parts - e^x integration
- Integration by specific formulaes - Formula 1
- Integration by specific formulaes - Formula 2
- Integration by specific formulaes - Formula 3
- Integration by specific formulaes - Formula 4
- Integration by specific formulaes - Formula 5
- Integration by specific formulaes - Formula 6
- Integration by specific formulaes - Formula 7
- Integration by specific formulaes - Formula 8
- Integration by specific formulaes - Method 9
- Integration by specific formulaes - Method 10
- Integration by partial fraction - Type 1
- Integration by partial fraction - Type 2
- Integration by partial fraction - Type 3
- Integration by partial fraction - Type 4
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Integration by partial fraction - Type 5
- Definite Integral as a limit of a sum
- Definite Integration - By Formulae
- Definite Integration - By Partial Fraction
- Definite Integration - By e formula
- Definite Integration - By Substitution
- Definite Integration by properties - P2
- Definite Integration by properties - P3
- Definite Integration by properties - P4
- Definite Integration by properties - P6
- Definite Integration by properties - P7
Transcript
Ex 7.5, 13 2 1 1 + 2 We can write the integrand as 2 1 1 + 2 = 2 1 1 + 2 = 2 1 1 + 2 Let 2 1 1 + 2 = 1 + + 1 + 2 2 1 1 + 2 = 1 + 2 + + 1 1 1 + 2 By cancelling denominator 2 = 1+ 2 + + 1 Putting x = 1 , in (1) 2 = A 1+ 1 2 + Bx + C 1 1 2 = A 1+1 + Bx+C 0 2 = 2A+0 2 2 = A 1 = A A = 1 Putting x = 0 , in (1) 2 = A 1+ 0 2 + B 0 +C 0 1 2 = A 1 +C 1 2 = A C Putting A = 1 2 = 1 C C = 1+2 C = 1 Now, Putting A = 1 , in (1) 2 = A 1+1 + B 1 +C 1 1 2 = 2 + + 2 2 = 2 +2 2 2 = 2+2 2 2+4=2 2 =2 =1 Thus, = 1, =1, = 1 So, we can write 2 1 1 + 2 = 1 + + 1 + 2 = 1 1 + + 1 2 + 1 Therefore integrating 2 1 1 + 2 = 1 1 + + 1 2 + 1 = 1 1 + 2 + 1 + 1 2 + 1 Solving 1 I2 = 2 + 1 Let = 2 +1 = 2 2 = Hence 2 + 1 = . 2 = 2 = 1 2 log + 1 Putting back t = 2 +1 = 1 2 log 2 +1 + 2 Therefore 2 1 1 + 2 = 1 1 + 2 + 1 + 1 2 + 1 = log 1 + 1 2 log 2 +1 + tan 1 + = + + + +
