Evaluate the following: d∫01xlog(1+2x) dx - Mathematics

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Sum

Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`

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Solution

Let I = `int_0^1 x log(1 + 2x)  "d"x`

= `[log (1 + 2x)  x^2/2]_0^1 - int_0^1  2/(1 + 2x)  x^2/2  "d"x`  .....[Integrating by parts]

= `1/2 [x^2 log (1 + 2x)]_0^1 - int_0^1  x^2/(1 + 2x)  "d"x`

= `1/2 [1 log 3 - 0] - int_0^1 (x/2 - x/(2(1 + 2x)))"d"x`

= ` 1/2 log 3 - 1/2 int_0^1 x "d"x + 1/2 int_0^1 x/(1 + 2x)  "d"x`

= `1/2 log 3 - 1/2 [x^2/2]_0^1 + 1/4 int_0^1  ((2x + 1 - 1))/((2x + 1))  "d"x`

= `1/2 log 3 - 1/2 [1/2 - 0] + 1/4 int_0^1 "d"x - 1/4 int_0^1  1/(1 + 2x)  "d"x`

= `1/2 log 3 - 1/4 + 1/4 - 1/8 [log (2x + 1)]_0^1`

= `1/2 log 3 - 1/4 + 1/4 - 1/8 [log 3 - log 1]`

= `1/2 log 3 - 1/8 log 3`

= `3/8 log 3`

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Chapter 7: Integrals - Exercise [Page 166]

APPEARS IN

NCERT Exemplar Mathematics Class 12
Chapter 7 Integrals
Exercise | Q 45 | Page 166

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