The function is measurable and negative on this interval. Its absolute value obeys

0 \leq \frac{- x l o g x}{( 1 + x ) ^{2}} \leq - x lo g x, \int_{0}^{1} - x lo g x d x = \frac{1}{4} .

x^{2} lo g x \to 0

at zero. The comparison test for integrals proves integrability.

b

Solution

For

x \geq 1

(1 + x)^{2} \leq 4 x^{2}

, so

\frac{x l o g x}{( 1 + x ) ^{2}} \geq \frac{l o g x}{4 x} \geq 0.

The integral of the lower bound from one to

R

(lo g R)^{2} /8

, which diverges. Thus the nonnegative function has infinite integral and is not integrable.

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