Ex 7.2, 35 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.2, 35 1 + log﷮𝑥﷯﷯﷮2﷯﷮𝑥﷯ Step 1: Let 1+ log﷮𝑥﷯= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0+ 1﷮𝑥﷯= 𝑑𝑡﷮𝑑𝑥﷯ 1﷮𝑥﷯= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 𝑥 . 𝑑𝑡 Step 2: Integrating the function ﷮﷮ 1 + log﷮𝑥﷯﷯﷮2﷯﷮𝑥﷯﷯. 𝑑𝑥 Putting 1− 𝑙𝑜𝑔﷮𝑥﷯=𝑡 & 𝑑𝑥=𝑥 . 𝑑𝑡 = ﷮﷮ 𝑡﷮2﷯﷮𝑥﷯ ﷯. 𝑥 . 𝑑𝑡 = ﷮﷮ 𝑡﷮2﷯﷯. 𝑑𝑡 = 𝑡﷮2 + 1﷯﷮2 + 1﷯ +𝐶 = 𝑡﷮3﷯﷮3﷯ +𝐶 = 𝟏﷮𝟑﷯ 𝟏+ 𝒍𝒐𝒈﷮𝒙﷯﷯﷮𝟑﷯+𝑪