Ex 7.2, 38 - Chapter 7 Class 12 Integrals
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 7.2, 38 10𝑥9+ 10𝑥 log𝑒10𝑥10+ 10𝑥 dx equals (A) 10𝑥 – 𝑥10 + 𝐶 (B) 10𝑥+ 𝑥10+𝐶 (C) 10𝑥 – 𝑥10−1 + 𝐶 (D) log 10𝑥+𝑥10 + 𝐶 Step 1: Let 𝑥10+ 10𝑥= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 10𝑥10−1+ 10𝑥 𝑙𝑜𝑔10= 𝑑𝑡𝑑𝑥 10𝑥9+ 10𝑥 𝑙𝑜𝑔10= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡 10𝑥9 + 10𝑥 𝑙𝑜𝑔10 Step 2: Integrating the function 10𝑥 910𝑥 𝑙𝑜𝑔10 𝑥10+ 10𝑥 . 𝑑𝑥 Putting 𝑥10+ 10𝑥=𝑡 & 𝑑𝑥= 𝑑𝑡 10𝑥9 + 10𝑥 𝑙𝑜𝑔10 = 10𝑥 910𝑥 𝑙𝑜𝑔10𝑡 . 𝑑𝑡 10𝑥9+ 10𝑥 𝑙𝑜𝑔10 = 1𝑡.𝑑𝑡 = log 𝑡+𝐶 = log 10𝑥+ 𝑥10+𝐶 ∴ Option D is correct.
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