The value of which of the following integrals is zero?

(A) $\int_0^1 x dx$

(B) $\int_{-1}^1 x dx$

(C) $\int_{-1}^1 x^2 dx$

(D) $\int_0^1 \log\left(\frac{x}{1-x}\right) dx$

Choose the correct answer from the options given below:

For option (A): $\int_0^1 x \, dx$

$\int_0^1 x \, dx = \frac{x^2}{2}\Bigg|_0^1$

$= \frac{1^2}{2} - \frac{0^2}{2}$

$= \frac{1}{2}$

This integral does not equal zero.

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For option (B): $\int_{-1}^1 x \, dx$

$\int_{-1}^1 x \, dx = \frac{x^2}{2}\Bigg|_{-1}^1$

$= \frac{1^2}{2} - \frac{(-1)^2}{2}$

$= \frac{1}{2} - \frac{1}{2}$

$= 0$

Alternatively, note that $f(x) = x$ is an odd function and the interval $[-1, 1]$ is symmetric about zero. The integral of an odd function over a symmetric interval is always zero, as the negative area on the left cancels the positive area on the right.

This integral equals zero.

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For option (C): $\int_{-1}^1 x^2 \, dx$

$\int_{-1}^1 x^2 \, dx = \frac{x^3}{3}\Bigg|_{-1}^1$

$= \frac{1^3}{3} - \frac{(-1)^3}{3}$

$= \frac{1}{3} - \left(\frac{-1}{3}\right)$

$= \frac{2}{3}$

Note that $x^2$ is an even function, not odd, so the areas add rather than cancel.

This integral does not equal zero.

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For option (D): $\int_0^1 \log\left(\frac{x}{1-x}\right) dx$

Let $I = \int_0^1 \log\left(\frac{x}{1-x}\right) dx$

Using the substitution $u = 1-x$, where $x = 1-u$ and $dx = -du$:

When $x = 0$, $u = 1$; when $x = 1$, $u = 0$

$I = \int_1^0 \log\left(\frac{1-u}{u}\right) (-du)$

$= \int_0^1 \log\left(\frac{1-u}{u}\right) du$

$= \int_0^1 \log\left(\frac{1-x}{x}\right) dx$

$= \int_0^1 \left[-\log\left(\frac{x}{1-x}\right)\right] dx$

$= -I$

Therefore: $I = -I$

$2I = 0$

$I = 0$

This integral equals zero.

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Both integrals (B) and (D) equal zero.