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{tex}\int {x\log 2xdx} {/tex}
{tex} = \int {\left( {\log 2x} \right)xdx} {/tex}
{tex}= \left( {\log 2x} \right)\int {xdx - \int {\left[ {\frac{d}{{dx}}\log 2x\int {xdx} } \right]dx} } {/tex}
[Applying product rule]
{tex}= \left( {\log 2x} \right)\frac{{{x^2}}}{2} - \int {\frac{1}{{2x}}.2.\frac{{{x^2}}}{2}dx} {/tex}
{tex} = \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\int {xdx} {/tex}
{tex}= \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\frac{{{x^2}}}{2} + c{/tex}
{tex}= \frac{{{x^2}}}{2}\log 2x - \frac{{{x^2}}}{4} + c{/tex}
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Preeti Dabral 3 years, 1 month ago
Given {tex}^a{\mu _g} = 1.5{/tex} and {tex}^a{\mu _w} = 1.3{/tex}
As, {tex}\delta = (\mu - 1)A{/tex}
For deviation in air, {tex}\mu = \frac{{{\mu _g}}}{{{\mu _a}}} = \frac{{1.5}}{1} = 1.50{/tex}
{tex}\therefore \delta = (1.5 - 1) \times 60^\circ{/tex} = 30°
For deviation in water, {tex}\mu = \frac{{{\mu _g}}}{{{\mu _w}}} = \frac{{1.5}}{{1.3}} = 1.15{/tex}
{tex}\therefore \delta = (1.15 - 1) \times 60^\circ {/tex} = 9°
Therefore, the angle of deviation is decreased.
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