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Misc 6 - Integrate 5x / (x + 1) (x2 + 9) - Class 12 CBSE - Integration by partial fraction - Type 5

Misc 6 - Chapter 7 Class 12 Integrals - Part 2
Misc 6 - Chapter 7 Class 12 Integrals - Part 3
Misc 6 - Chapter 7 Class 12 Integrals - Part 4
Misc 6 - Chapter 7 Class 12 Integrals - Part 5
Misc 6 - Chapter 7 Class 12 Integrals - Part 6
Misc 6 - Chapter 7 Class 12 Integrals - Part 7
Misc 6 - Chapter 7 Class 12 Integrals - Part 8

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Transcript

Misc 6 Integrate the function 5𝑥﷮ 𝑥 + 1﷯ 𝑥﷮2﷯ + 9﷯﷯ Let I = ﷮﷮ 5𝑥﷮ 𝑥 + 1﷯ 𝑥﷮2﷯ + 9﷯﷯𝑑𝑥﷯ We can write integrate as : 5𝑥﷮ 𝑥 + 1﷯ 𝑥﷮2﷯+ 9﷯﷯= A﷮𝑥 + 1﷯+ B𝑥 + 𝐶﷮ 𝑥﷮2﷯ + 9﷯ By cancelling denominators 5𝑥=A 𝑥﷮2﷯+9﷯+ B𝑥+𝐶﷯ 𝑥+1﷯ Putting 𝑥=−1 5 × −1﷯=A −1﷯﷮2﷯+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 0﷮2﷯+9﷯+ B ×0+𝐶﷯ 0+1﷯ 0 = A 9﷯+ 0+𝐶﷯ 1﷯ 0 = 9A + C C = – 9A C = – 9 × −1﷮ 2﷯ C = 9﷮2﷯ Putting x = 1 5(1)= A 1﷮2﷯+9﷯+ B1+𝐶﷯ 1+1﷯ 5(1)= A(10) +(B + C) (2) Putting A = −1﷮2﷯ and C = 9﷮2﷯ 5(1) = −1﷮2﷯﷯ (10) + 2B + 2C 5(1) = −5 + 2B + 2 9﷮2﷯﷯ 5(1) = −5 + 2B + 9 5 = 2B + 4 1 = 2B 1﷮2﷯ = B Hence we can write 5𝑥﷮ 𝑥+1﷯ 𝑥﷮2﷯+9﷯﷯= −1﷮ 2﷯﷯﷮𝑥 + 1﷯+ 1﷮2﷯ 𝑥 + 9﷮2﷯﷮ 𝑥﷮2﷯ + 9﷯ = −1﷮2 𝑥 + 1﷯﷯+ 𝑥﷮2 𝑥﷮2﷯ + 9﷯﷯+ 9﷮2 𝑥﷮2﷯ + 9﷯﷯ Integrating w.r.t.𝑥 ﷮﷮ 5𝑥﷮ 𝑥 + 1﷯ 𝑥﷮2﷯ + 9﷯﷯𝑑𝑥= ﷮﷮ −1﷮2 𝑥 + 1﷯﷯+ 𝑥﷮2 𝑥﷮2﷯ + 9﷯﷯+ 9﷮2 𝑥﷮2﷯ + 9﷯﷯﷯﷯﷯ = ﷮﷮ −1﷮2 𝑥 + 1﷯﷯𝑑𝑥+﷯ ﷮﷮ 𝑥﷮2 𝑥﷮2﷯ + 9﷯﷯𝑑𝑥+﷯ ﷮﷮ 9﷮2 𝑥﷮2﷯ + 9﷯﷯﷯𝑑𝑥 Hence I = I1 + I2 + I3 I1 = ﷮﷮ −1﷮2 𝑥 + 1﷯﷯𝑑𝑥﷯ = −1﷮ 2 ﷯𝑙𝑜𝑔 𝑥+1﷯+𝐶1 I2 = ﷮﷮ 𝑥﷮2 𝑥﷮2﷯ + 9﷯﷯𝑑𝑥﷯ = 1﷮2﷯ ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 9﷯𝑑𝑥﷯ Putting 𝑡= 𝑥﷮2﷯+9 Differentiating w.r.t. 𝑥 𝑑𝑡﷮𝑑𝑥﷯=2𝑥 𝑑𝑡﷮2𝑥﷯=𝑑𝑥 Therefore 1﷮2﷯ ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 9﷯𝑑𝑥﷯= 1﷮2﷯ ﷮﷮ 𝑥﷮𝑡﷯ × 𝑑𝑡﷮2𝑥﷯﷯ = 1﷮2﷯ ﷮﷮ 𝑑𝑡﷮2𝑡﷯﷯ = 1﷮2 ×2﷯ ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ = 1﷮4﷯log 𝑡﷯+𝐶2 Putting back 𝑡= 𝑥﷮2﷯+9 = 1﷮4﷯𝑙𝑜𝑔 𝑥﷮2﷯+9﷯+𝐶2 Now, I3 = ﷮﷮ 9﷮2 𝑥﷮2﷯ + 9﷯﷯﷯𝑑𝑥 = 9﷮2﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 9﷯𝑑𝑥﷯ Now, I3 = ﷮﷮ 9﷮2 𝑥﷮2﷯ + 9﷯﷯﷯𝑑𝑥 = 9﷮2﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 9﷯𝑑𝑥﷯ = 9﷮2﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 3﷮2﷯﷯𝑑𝑥﷯ = 9﷮2﷯ × 1﷮3﷯ tan﷮−1﷯﷮ 𝑥﷮3﷯+𝐶3﷯ = 3﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮3﷯+﷯ 𝐶3 Hence I = I1 + I2 + I3 = −1﷮ 2﷯𝑙𝑜𝑔 𝑥+1﷯+𝐶1+ 1﷮4﷯𝑙𝑜𝑔 𝑥﷮2﷯+9﷯+𝐶2+ 3﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮3﷯+𝐶3﷯ = −1﷮ 2﷯𝑙𝑜𝑔 𝑥+1﷯+ 1﷮4﷯𝑙𝑜𝑔 𝑥﷮2﷯+9﷯+ 3﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮3﷯+𝐶1+𝐶2+𝐶3﷯ = −𝟏﷮ 𝟐﷯𝒍𝒐𝒈 𝒙+𝟏﷯+ 𝟏﷮𝟒﷯𝒍𝒐𝒈 𝒙﷮𝟐﷯+𝟗﷯+ 𝟑﷮𝟐﷯ 𝐭𝐚𝐧﷮−𝟏﷯﷮ 𝒙﷮𝟑﷯+𝑪﷯

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.