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Question
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Solution
We have I = `int sqrt(10 - 4x + 4x^2) "d"x`
= `int sqrt((2x - 1)^2 + (3)^2) "d"x`
Put t = 2x – 1
Then dt = 2dx.
Therefore, I = `1/2 int sqrt("t"^2 + (3)^2) "dt"`
= `1/2 "t" sqrt("t"^2 + 9)/2 + 9/4 log|"t" + sqrt("t"^2 + 9)| + "C"`
= `1/4(2x - 1) sqrt((2x - 1)^2 + 9) + 9/4 log|(2x - 1) + sqrt((2x - 1)^2 + 9)| + "C"`
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