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Question
Evaluate:
∫ (log x)2 dx
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Solution
Let I = ∫ (log x)2 dx
I = ∫ (log x)2 . 1 dx
I = `(log x)^2 int 1. "dx" - int ["d"/"dx" (log x)^2 int 1. "dx"] "dx"`
I = `x(log x)^2 - int 2 log x. 1/cancelx. cancelx "dx"`
I = `x(log x)^2 - 2 int log x. 1 "dx"`
I = `x(log x)^2 - 2[log x int 1. "dx" - int {"d"/"dx" (log x) int 1. "dx"}]`dx
I = `x(log x)^2 - 2[(log x)x - int 1/cancelx. cancelx. "dx"]`
I = `x(log x)^2 - 2[xlog x - int 1. "dx"]`
I = x(log x)2 – 2(x log x – x) + c
∴ I = x(log x)2 – 2x log x + 2x + c
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