Chapter 7 Class 12 Integrals

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Example 10 (i) - Chapter 7 Class 12 Integrals (Important Question)

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Example 10 - Find integrals (i) x + 2 / 2x2 + 6x + 5 dx - Examples

Example 10 - Chapter 7 Class 12 Integrals - Part 2
Example 10 - Chapter 7 Class 12 Integrals - Part 3 Example 10 - Chapter 7 Class 12 Integrals - Part 4 Example 10 - Chapter 7 Class 12 Integrals - Part 5 Example 10 - Chapter 7 Class 12 Integrals - Part 6 Example 10 - Chapter 7 Class 12 Integrals - Part 7

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Example 10 Find the following integrals: (i) āˆ«1ā–’(š‘„ + 2)/(2š‘„^2 + 6š‘„ + 5 ) š‘‘š‘„ We can write numerator as š‘„+2= A š‘‘/š‘‘š‘„ (2š‘„^2+6š‘„+5) + B š‘„+2= A [4š‘„+6]+ B š‘„+2=4š“š‘„+6A+B Finding A & B Comparing coefficient of š‘„ š‘„=4š“š‘„ 1 =4A A=1/4 Comparing constant term 2=6A+B 2=6(1/4)+B 2=3/2+B B=2āˆ’3/2=1/2 Now, we know that š‘„+2= A [4š‘„+6]+ B š‘„+2=1/4 [4š‘„+6]+1/2 Now, our equation is āˆ«1ā–’怖(š‘„ + 2)/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„=āˆ«1ā–’怖(1/4 [4š‘„ + 6] + 1/2)/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„怗怗 =āˆ«1ā–’怖(1/4 [4š‘„ + 6])/(2š‘„^2 + 6š‘„ + 5)+āˆ«1ā–’怖(1/2)/(2š‘„^2+6š‘„+5).š‘‘š‘„怗怗 =1/4 āˆ«1ā–’怖(4š‘„ + 6)/(2š‘„^2 + 6š‘„ + 5) š‘‘š‘„+1/2 āˆ«1ā–’怖1/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„怗怗 Solving I1 I1 =1/4 āˆ«1ā–’怖(4š‘„ + 6)/(2š‘„^2 + 6š‘„ + 5) š‘‘š‘„怗 Let t = 2š‘„^2 + 6š‘„ + 5 Differentiating both sides w.r.t.š‘„ 4š‘„ +6=š‘‘š‘”/š‘‘š‘„ š‘‘š‘„=š‘‘š‘”/(4š‘„ + 6) Now, I1 =1/4 āˆ«1ā–’怖(4š‘„ + 6)/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„怗 Putting the values of (2š‘„^2+6š‘„+5) and š‘‘š‘„, we get I1 =1/4 āˆ«1ā–’怖(4š‘„ + 6)/š‘”.š‘‘š‘”/(4š‘„ + 6) 怗 I1 =1/4 āˆ«1ā–’怖1/š‘”.š‘‘š‘” 怗 I1 =1/4 š‘™š‘œš‘”|š‘”|+š¶1 I1 =1/4 š‘™š‘œš‘”|2š‘„^2+6š‘„+5|+š¶1 Solving I2 I2 =1/2 āˆ«1ā–’怖1/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„ 怗 I2 =1/2 āˆ«1ā–’怖1/2[š‘„^2 + 6š‘„/2 + 5/2 ] .š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/(š‘„^2 +3š‘„ + 5/2).š‘‘š‘„ 怗 (Using āˆ«1ā–’怖1/š‘„.š‘‘š‘„=š‘™š‘œš‘”|š‘„|+š¶1怗) (Using š‘”=2š‘„^2+6š‘„+5) I2 =1/4 āˆ«1ā–’怖1/(š‘„^2 + 2(š‘„)(3/2) + 5/2).š‘‘š‘„ 怗 Adding & subtracting (3/2)^2 in denominator I2 =1/4 āˆ«1ā–’怖1/(š‘„^2 + 2(š‘„) 怖(3/2) +(3/2)^2āˆ’ (3/2)怗^2 + 5/2).š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/((š‘„ + 3/2)^2 āˆ’ (3/2)^2 + 5/2) š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/((š‘„ + 3/2)^2 āˆ’ 9/4 + 5/2).š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/((š‘„ + 3/2)^2+ (āˆ’9 + 10)/4 ).š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/((š‘„ + 3/2)^2+ 1/4 ).š‘‘š‘„ 怗 I2 =1/4 āˆ«1ā–’怖1/((š‘„ + 3/2)^2+ (1/2)^2 ).š‘‘š‘„ 怗 =1/4 [1/(1/2) tan^(āˆ’1)ā”怖(š‘„ + 3/2)/(1/2)+š¶2怗 ] =1/4 [2 tan^(āˆ’1)ā”怖((2š‘„ + 3)/2)/(1/2)+š¶2怗 ] =1/4 [2 怖tan^(āˆ’1) (2š‘„+3)怗ā”怖+š¶2怗 ] It is of form āˆ«1ā–’怖š‘‘š‘„/(š‘„^2 + š‘Ž^2 )=1/š‘Ž 怖š‘”š‘Žš‘›ć€—^(āˆ’1)ā”怖š‘„/š‘Ž+š¶2" " 怗 怗 Replacing š‘„ by (š‘„+3/2) and by 1/2, we get =2/4 tan^(āˆ’1)ā”怖(2š‘„+3)+š¶2/4怗 =1/2 tan^(āˆ’1)ā”怖(2š‘„+3)+š¶3怗 Now, putting the value of I1 and I2 in eq. (1) āˆ“ āˆ«1ā–’怖(š‘„+2)/(2š‘„^2 + 6š‘„ + 5).š‘‘š‘„怗 =1/4 š‘™š‘œš‘”|2š‘„^2+6š‘„+5|+š¶1+1/2 tan^(āˆ’1)ā”怖(2š‘„+3)+怗 š¶3 =šŸ/šŸ’ š’š’š’ˆ|šŸš’™^šŸ+šŸ”š’™+šŸ“|+šŸ/šŸ 怖š’•š’‚š’ć€—^(āˆ’šŸ)ā”怖(šŸš’™+šŸ‘)+怗 š‘Ŗ (where š¶3 = š¶2/4 )

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