Ex 7.2, 3 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Ex 7.2, 3 Integrate the function: 1/(𝑥 + 𝑥 log𝑥 ) 1/(𝑥 + 𝑥 log𝑥 )=1/𝑥(1 + log𝑥 ) Step 1: Let 1+log𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0 + 1/𝑥=𝑑𝑡/𝑑𝑥 1/𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑥 𝑑𝑡 Step 2: Integrating function ∫1▒1/(𝑥 + 𝑥 log𝑥 ) .𝑑𝑥 =∫1▒1/(𝑥 (1 + log𝑥 ) ) .𝑑𝑥" " Putting 1+log𝑥 & 𝑑𝑥=𝑥 𝑑𝑡 = ∫1▒1/(𝑥(𝑡)) 𝑑𝑡.𝑥 = ∫1▒1/𝑡 𝑑𝑡 = 𝑙𝑜𝑔|𝑡|+𝐶 Putting back 𝑡 =1+𝑙𝑜𝑔𝑥 = 𝒍𝒐𝒈|𝟏+𝐥𝐨𝐠𝒙 |+𝑪 (𝑈𝑠𝑖𝑛𝑔∫1▒〖1/𝑥 𝑑𝑥〗=log|𝑥|+𝐶)
