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Ex 7.2, 38 - Integration 10x9 + 10x loge 10 / x10 + 10x - Integration by substitution - lnx

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.2, 38 ﷮﷮ 10𝑥﷮9﷯+ 10﷮𝑥﷯ log﷮𝑒﷯﷮10﷯﷮𝑥10+ 10𝑥﷯﷯ dx equals (A) 10𝑥 – 𝑥﷮10﷯ + 𝐶 (B) 10𝑥+ 𝑥﷮10﷯+𝐶 (C) 10𝑥 – 𝑥﷮10﷯﷯﷮−1﷯ + 𝐶 (D) log﷮ 10𝑥+𝑥10﷯﷯ + 𝐶 Step 1: Let 𝑥10+ 10𝑥= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 10𝑥﷮10−1﷯+ 10﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯= 𝑑𝑡﷮𝑑𝑥﷯ 10𝑥﷮9﷯+ 10﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮ 10𝑥﷮9﷯ + 10﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯﷯ Step 2: Integrating the function ﷮﷮ 10𝑥 910﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯﷮ 𝑥﷮10﷯+ 10﷮𝑥﷯﷯﷯ . 𝑑𝑥 Putting 𝑥﷮10﷯+ 10﷮𝑥﷯﷯=𝑡 & 𝑑𝑥= 𝑑𝑡﷮ 10𝑥﷮9﷯ + 10﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯﷯ = ﷮﷮ 10𝑥 910﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯﷮𝑡﷯﷯ . 𝑑𝑡﷮ 10𝑥﷮9﷯+ 10﷮𝑥﷯ 𝑙𝑜𝑔﷮10﷯﷯ = ﷮﷮ 1﷮𝑡﷯﷯.𝑑𝑡 = log 𝑡﷯+𝐶 = log 10﷮𝑥﷯+ 𝑥﷮10﷯﷯+𝐶 ∴ Option D is correct.

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