# Misc 6 - Chapter 7 Class 12 Integrals

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 6 Integrate the function 5𝑥 𝑥 + 1 𝑥2 + 9 Let I = 5𝑥 𝑥 + 1 𝑥2 + 9𝑑𝑥 We can write integrate as : 5𝑥 𝑥 + 1 𝑥2+ 9= A𝑥 + 1+ B𝑥 + 𝐶 𝑥2 + 9 By cancelling denominators 5𝑥=A 𝑥2+9+ B𝑥+𝐶 𝑥+1 Putting 𝑥=−1 5 × −1=A −12+9+ B × −1+𝐶 −1+1 – 5 = A 1+9+ −B+𝐶 0 – 5 = A × 10 – 5 = 10A A = −5 10 A = −1 2 Similarly putting 𝑥=0 5 × 0 = A 02+9+ B ×0+𝐶 0+1 0 = A 9+ 0+𝐶 1 0 = 9A + C C = – 9A C = – 9 × −1 2 C = 92 Putting x = 1 5(1)= A 12+9+ B1+𝐶 1+1 5(1)= A(10) +(B + C) (2) Putting A = −12 and C = 92 5(1) = −12 (10) + 2B + 2C 5(1) = −5 + 2B + 2 92 5(1) = −5 + 2B + 9 5 = 2B + 4 1 = 2B 12 = B Hence we can write 5𝑥 𝑥+1 𝑥2+9= −1 2𝑥 + 1+ 12 𝑥 + 92 𝑥2 + 9 = −12 𝑥 + 1+ 𝑥2 𝑥2 + 9+ 92 𝑥2 + 9 Integrating w.r.t.𝑥 5𝑥 𝑥 + 1 𝑥2 + 9𝑑𝑥= −12 𝑥 + 1+ 𝑥2 𝑥2 + 9+ 92 𝑥2 + 9 = −12 𝑥 + 1𝑑𝑥+ 𝑥2 𝑥2 + 9𝑑𝑥+ 92 𝑥2 + 9𝑑𝑥 Hence I = I1 + I2 + I3 I1 = −12 𝑥 + 1𝑑𝑥 = −1 2 𝑙𝑜𝑔 𝑥+1+𝐶1 I2 = 𝑥2 𝑥2 + 9𝑑𝑥 = 12 𝑥 𝑥2 + 9𝑑𝑥 Putting 𝑡= 𝑥2+9 Differentiating w.r.t. 𝑥 𝑑𝑡𝑑𝑥=2𝑥 𝑑𝑡2𝑥=𝑑𝑥 Therefore 12 𝑥 𝑥2 + 9𝑑𝑥= 12 𝑥𝑡 × 𝑑𝑡2𝑥 = 12 𝑑𝑡2𝑡 = 12 ×2 𝑑𝑡𝑡 = 14log 𝑡+𝐶2 Putting back 𝑡= 𝑥2+9 = 14𝑙𝑜𝑔 𝑥2+9+𝐶2 Now, I3 = 92 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 9𝑑𝑥 Now, I3 = 92 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 32𝑑𝑥 = 92 × 13 tan−1 𝑥3+𝐶3 = 32 tan−1 𝑥3+ 𝐶3 Hence I = I1 + I2 + I3 = −1 2𝑙𝑜𝑔 𝑥+1+𝐶1+ 14𝑙𝑜𝑔 𝑥2+9+𝐶2+ 32 tan−1 𝑥3+𝐶3 = −1 2𝑙𝑜𝑔 𝑥+1+ 14𝑙𝑜𝑔 𝑥2+9+ 32 tan−1 𝑥3+𝐶1+𝐶2+𝐶3 = −𝟏 𝟐𝒍𝒐𝒈 𝒙+𝟏+ 𝟏𝟒𝒍𝒐𝒈 𝒙𝟐+𝟗+ 𝟑𝟐 𝐭𝐚𝐧−𝟏 𝒙𝟑+𝑪

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5

Misc 6 You are here

Misc 7

Misc 8 Important

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21

Misc 22

Misc 23

Misc 24 Important

Misc 25

Misc 26

Misc 27

Misc 28

Misc 29

Misc 30 Important

Misc 31

Misc 32 Important

Misc 33

Misc 34

Misc 35

Misc 36

Misc 37

Misc 38

Misc 39

Misc 40

Misc 41 Important

Misc 42

Misc 43

Misc 44 Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.