Ex 7.2, 3 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.2, 3 Integrate the function: 1﷮𝑥 + 𝑥 log﷮𝑥﷯﷯ 1﷮𝑥 + 𝑥 log﷮𝑥﷯﷯= 1﷮𝑥 1 + log﷮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 function ﷮﷮ 1﷮𝑥 + 𝑥 log﷮𝑥﷯﷯﷯ .𝑑𝑥 = ﷮﷮ 1﷮𝑥 1 + log﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 Putting 1+ log﷮𝑥﷯ & 𝑑𝑥=𝑥 𝑑𝑡 = ﷮﷮ 1﷮𝑥(𝑡)﷯﷯ 𝑑𝑡.𝑥 = ﷮﷮ 1﷮(𝑡)﷯﷯ 𝑑𝑡 = 𝑙𝑜𝑔 𝑡﷯+𝐶 putting 𝑡 =1+ 𝑙𝑜𝑔﷮𝑥﷯ = 𝒍𝒐𝒈 𝟏+ 𝐥𝐨𝐠﷮𝒙﷯﷯+𝑪