Ex 7.2, 3
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 7.2, 3 Integrate the function: 1𝑥 + 𝑥 log𝑥 1𝑥 + 𝑥 log𝑥= 1𝑥 1 + log𝑥 Step 1: Let 1+ log𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0 + 1𝑥= 𝑑𝑡𝑑𝑥 1𝑥= 𝑑𝑡𝑑𝑥 𝑑𝑥=𝑥 𝑑𝑡 Step 2: Integrating function 1𝑥 + 𝑥 log𝑥 .𝑑𝑥 = 1𝑥 1 + log𝑥 .𝑑𝑥 Putting 1+ log𝑥 & 𝑑𝑥=𝑥 𝑑𝑡 = 1𝑥(𝑡) 𝑑𝑡.𝑥 = 1(𝑡) 𝑑𝑡 = 𝑙𝑜𝑔 𝑡+𝐶 putting 𝑡 =1+ 𝑙𝑜𝑔𝑥 = 𝒍𝒐𝒈 𝟏+ 𝐥𝐨𝐠𝒙+𝑪
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