MCQ
If $\int_0^1 x \log \left(1+\frac{x}{2}\right) d x= a + b \log \frac{2}{3}$, then
- A$a =\frac{3}{2}, b=\frac{3}{2}$
- B$a=\frac{3}{4}, b=-\frac{3}{4}$
- ✓$a =\frac{3}{4}, b=\frac{3}{2}$
- D$a=b$
✓
Answer
Correct option: C.
$a =\frac{3}{4}, b=\frac{3}{2}$
(C)
$\int_0^1 x \log \left(1+\frac{x}{2}\right) d x$
$=\left[\log \left(1+\frac{x}{2}\right) \cdot \frac{x^2}{2}\right]_0^1-\int_0^1\left[\frac{1}{1+\frac{x}{2}} \cdot \frac{1}{2} \cdot \frac{x^2}{2}\right] d x$
$=\frac{1}{2} \log \frac{3}{2}-\frac{1}{2} \int_0^1 \frac{x^2}{x+2} d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{2} \int_0^1\left[x-\frac{2 x}{x+2}\right] d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{2}\left[\frac{x^2}{2}\right]_0^1+\int_0^1 \frac{x}{x+2} d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+\int_0^1\left[1-\frac{2}{x+2}\right] d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+[x-2 \log (x+2)]_0^1 \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+1-2 \log 3+2 \log 2 \\ =\frac{3}{4}+\frac{3}{2} \log \frac{2}{3} \\ \therefore a =\frac{3}{4}, b=\frac{3}{2}$
$\int_0^1 x \log \left(1+\frac{x}{2}\right) d x$
$=\left[\log \left(1+\frac{x}{2}\right) \cdot \frac{x^2}{2}\right]_0^1-\int_0^1\left[\frac{1}{1+\frac{x}{2}} \cdot \frac{1}{2} \cdot \frac{x^2}{2}\right] d x$
$=\frac{1}{2} \log \frac{3}{2}-\frac{1}{2} \int_0^1 \frac{x^2}{x+2} d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{2} \int_0^1\left[x-\frac{2 x}{x+2}\right] d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{2}\left[\frac{x^2}{2}\right]_0^1+\int_0^1 \frac{x}{x+2} d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+\int_0^1\left[1-\frac{2}{x+2}\right] d x \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+[x-2 \log (x+2)]_0^1 \\ =\frac{1}{2} \log \frac{3}{2}-\frac{1}{4}+1-2 \log 3+2 \log 2 \\ =\frac{3}{4}+\frac{3}{2} \log \frac{2}{3} \\ \therefore a =\frac{3}{4}, b=\frac{3}{2}$
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