If int_0^1 {x log left( {1 + frac{x}{2}} righ | vedclass

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(C) To evaluate the integral $I = \int_0^1 x \log \left( 1 + \frac{x}{2} \right) dx$,we use integration by parts: $\int u v dx = u \int v dx - \int (u

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$a = \frac{3}{4}, b = \frac{3}{2}$

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(C) To evaluate the integral $I = \int_0^1 x \log \left( 1 + \frac{x}{2} \right) dx$,we use integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.Let $u = \log \left( 1 + \frac{x}{2} \right)$ and $v = x$.Then $du = \frac{1}{1 + x/2} \cdot \frac{1}{2} dx = \frac{1}{x+2} dx$ and $\int v dx = \frac{x^2}{2}$.$I = \left[ \frac{x^2}{2} \log \left( 1 + \frac{x}{2} \right) \right]_0^1 - \int_0^1 \frac{x^2}{2(x+2)} dx$$I = \left( \frac{1}{2} \log \frac{3}{2} - 0 \right) - \frac{1}{2} \int_0^1 \frac{x^2}{x+2} dx$Using polynomial division,$\frac{x^2}{x+2} = x - 2 + \frac{4}{x+2}$.$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \int_0^1 \left( x - 2 + \frac{4}{x+2} \right) dx$$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left[ \frac{x^2}{2} - 2x + 4 \log(x+2) \right]_0^1$$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left( \left( \frac{1}{2} - 2 + 4 \log 3 \right) - (0 - 0 + 4 \log 2) \right)$$I = \frac{1}{2} \log \frac{3}{2} - \frac{1}{2} \left( -\frac{3}{2} + 4 \log \frac{3}{2} \right)$$I = \frac{1}{2} \log \frac{3}{2} + \frac{3}{4} - 2 \log \frac{3}{2} = \frac{3}{4} - \frac{3}{2} \log \frac{3}{2} = \frac{3}{4} + \frac{3}{2} \log \frac{2}{3}$.Comparing with $a + b \log \frac{2}{3}$,we get $a = \frac{3}{4}$ and $b = \frac{3}{2}$.

If $\int_0^1 {x \log \left( {1 + \frac{x}{2}} \right)} \,dx = a + b \log \frac{2}{3},$ then

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