Ex 7.2, 35
Last updated at March 11, 2017 by Teachoo
Transcript
Ex 7.2, 35 1 + log𝑥2𝑥 Step 1: Let 1+ log𝑥= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0+ 1𝑥= 𝑑𝑡𝑑𝑥 1𝑥= 𝑑𝑡𝑑𝑥 𝑑𝑥 = 𝑥 . 𝑑𝑡 Step 2: Integrating the function 1 + log𝑥2𝑥. 𝑑𝑥 Putting 1− 𝑙𝑜𝑔𝑥=𝑡 & 𝑑𝑥=𝑥 . 𝑑𝑡 = 𝑡2𝑥 . 𝑥 . 𝑑𝑡 = 𝑡2. 𝑑𝑡 = 𝑡2 + 12 + 1 +𝐶 = 𝑡33 +𝐶 = 𝟏𝟑 𝟏+ 𝒍𝒐𝒈𝒙𝟑+𝑪
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