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Ex 7.7, 10 - Chapter 7 Class 12 Integrals
Last updated at July 14, 2026 by Teachoo
Integration by specific formulaes - Formula 7
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
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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
- 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
Ex7.7, 10 1+ 2 dx is equal to 2 1+ 2 + 1 2 log + 1+ 2 + C 2 2 (1+ 2) 2 3 + C 2 2 x (1+ 2) 3 2 + C 2 2 1+ 2 + 1 2 x2 log + 1+ 2 + C 1+ 2 = 2 + 1 2 = 2 2 + 1 2 + 1 2 2 + 2 + 1 2 + = 2 1+ 2 + 1 2 + 2 +1 + Option A is correct.
