# Misc 5

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 5 Integrate the function 1 𝑥 12 + 𝑥 13 1 𝑥 12 + 𝑥 13 𝑑𝑥 Let x = 𝑡6 𝑑𝑥𝑑𝑡=6 𝑡5 dx = 6 𝑡5 dt Substituting value of x and dx = 6𝑡5 (𝑡6) 12 + (𝑡6) 13 = 6𝑡5 𝑡3 + 𝑡2 𝑑𝑡 = 6 𝑡3𝑡 + 1 𝑑𝑡 Adding and subtracting 1 in numerator = 6 𝑡3 + 1 − 1𝑡 + 1 𝑑𝑡 = 6 𝑡3 + 1 𝑡 + 1− 1𝑡 + 1 𝑑𝑡 = 6 𝑡 + 1( 𝑡2+ 1 − 𝑡)𝑡 + 1− 1𝑡 + 1 𝑑𝑡 = 𝑡2+1−𝑡− 11 + 𝑡 𝑑𝑡 = 6 𝑡33+𝑡− 𝑡22−𝑙𝑜𝑔 1+𝑡+ C = 2 𝑡3+6𝑡− 3𝑡2−6 𝑙𝑜𝑔 1+𝑡+ C Putting back value of t = 2 (𝑥 16)3+6 (𝑥 16)−3 (𝑥 16)2−6 𝑙𝑜𝑔 1+ 𝑥 16+ C = 2 𝑥 12+6 𝑥 16−3 𝑥 13−6 𝑙𝑜𝑔 1+ 𝑥 16+ C = 2 𝑥−3 𝑥 13+ 6 𝑥 16−6 log(1+ 𝑥 16)+ C

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Misc 3

Misc 4

Misc 5 You are here

Misc 6

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Misc 8 Important

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Misc 11

Misc 12

Misc 13

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Misc 16

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Misc 18 Important

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Misc 24 Important

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Misc 30 Important

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Misc 32 Important

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Misc 41 Important

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Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.