∫ 1 0 X Log ( 1 + 2 X ) D X - Mathematics

Question

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

Sum

Solution

\[\text{Let I }=\int_0^1 x\log\left( 1 + 2x \right)dx\]

Applying integration by parts, we have

\[I = \log\left( 1 + 2x \right)\frac{x^2}{2}_0^1 - \int_0^1 \left( \frac{2}{1 + 2x} \right) \times \frac{x^2}{2}dx\]
\[ = \frac{1}{2}\left( \log3 - 0 \right) - \int_0^1 \frac{x^2}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{4 x^2 - 1 + 1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{\left( 2x + 1 \right)\left( 2x - 1 \right)}{1 + 2x}dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \left( 2x - 1 \right)dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]

\[= \left.\frac{1}{2}\log3 - \frac{1}{4} \times \frac{\left( 2x - 1 \right)^2}{2 \times 2}\right|_0^1 - \left.\frac{1}{4} \times \frac{\log\left( 1 + 2x \right)}{2}\right|_0^1 \]
\[ = \frac{1}{2}\log3 - \frac{1}{16}\left( 1 - 1 \right) - \frac{1}{8}\left( \log3 - \log1 \right)\]
\[ = \frac{1}{2}\log3 - 0 - \frac{1}{8}\log3 ....................\left( \log1 = 0 \right)\]
\[ = \frac{3}{8}\log3\]

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Definite Integrals

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Q 65 Q 64 Q 66

Chapter 20: Definite Integrals - Exercise 20.1 [Page 18]

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RD Sharma Mathematics [English] Class 12

Chapter 20 Definite Integrals
Exercise 20.1 | Q 65 | Page 18

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