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Ex 7.5, 19 - Chapter 7 Class 12 Integrals
Last updated at July 14, 2026 by Teachoo
Integration by partial fraction - Type 1
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
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Integration by partial fraction - Type 1
- Integration by partial fraction - Type 2
- Integration by partial fraction - Type 3
- Integration by partial fraction - Type 4
- 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
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Transcript
Ex 7.5, 19 2 ( 2 + 1)( 2 + 3) Let 2 = Differentiate both sides . . . . 2 = = 2 Substituting value of & 2 ( 2 + 1)( 2 + 3) = 2 + 1 + 3 = 2 + 1 + 3 2 = + 1 + 3 Now we can write 1 + 1 + 3 = + 1 + + 3 1 + 1 + 3 = + 3 + + 1 + 1 + 3 Cancelling denominator 1 = +3 + +1 Putting t = 1 in (1) 1 = 1+3 + 1+1 1 = 2+ 0 1 = 2 = 1 2 Similarly Putting t= 3 in (1) 1 = +3 + B ( +1) 1 = 3+3 + 3+1 1 = 0+ 2 1 = 2 = 1 2 Therefore 1 + 1 + 3 = 1 2 + 1 + 1 2 + 3 = 1 2 log +1 1 2 log +3 + = 1 2 log [ +1 log |t+3|]+C = 1 2 log + 1 + 3 + Substituting back the value of t = 2 = 1 2 log 2 + 1 2 + 3 + = + + +
