Question

If $I={\int }_0^1\frac{xdx}{8+x^3}$ then the smallest interval in which $I$ lies is

• A. $(0,\frac{1}{8})$
• B. $(0,\frac{1}{9})$
• C. $(0,\frac{1}{10})$
• D. $(0,\frac{1}{7})$

Answer 1

Answer: (B)

$(0,\frac{1}{9})$

$I={\int }_0^1\frac{x}{8+x^3}dx={\int }_0^1\frac{x}{(x+2)(x^2-2x+4)}dx$
$={\int }_0^1(\frac{x+2}{6(x^2-2x+4)}-\frac{1}{6(x+2)})dx$
$=\frac{1}{6}{\int }_0^1(\frac{2x-2}{2(x^2-2x+4)}+\frac{3}{6x^2-2x+4})dx-\frac{1}{6}\int \frac{1}{x+2}dx$
$=\frac{1}{2}{\int }_0^1\frac{2x-2}{x^2-2x+4}dx+\frac{1}{2}{\int }_0^1\frac{1}{x^2-2x+4}-\frac{1}{6}{\int }_0^1\frac{1}{x+2}dx$
$=[\frac{1}{12}\log (x^2-2x+4)-\frac{1}{6}\log (x+2)]+\frac{{\tan }^{-1}\left(\frac{x-1}{\sqrt{3}}\right)}{2\sqrt{3}}$
$=\frac{1}{12}\log \left(3\right)-\frac{1}{6}\log \left(3\right)+\frac{{\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)}{2\sqrt{3}}$
$-\frac{1}{12}\log \left(4\right)-\frac{1}{6}\log \left(2\right)+\frac{{\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)}{2\sqrt{3}}$
And this lies between $(0,\frac{1}{9})$