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Find d∫10-4x+4x2 dx

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Question

Find `int sqrt(10 - 4x + 4x^2)  "d"x`

Sum
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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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Chapter 7: Integrals - Solved Examples [Page 153]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 7 Integrals
Solved Examples | Q 14 | Page 153

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