What is the integral from 0 to 1 of xln(1+1/x) ?

The integral from 0 to 1 of xln(1+1/x) is 1/2

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integral from 0 to 1 of xln(1+1/x)

\int_{0}^{1} x ln (1 + \frac{1}{x}) d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{1} x ln (1 + \frac{1}{x}) d x

Apply Integration By Parts

= [\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - \int - \frac{x}{2 ( x + 1 )} d x]_{0}^{1}

\int - \frac{x}{2 ( x + 1 )} d x = - \frac{1}{2} (x + 1 - ln ∣ x + 1 ∣)

= [\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - (- \frac{1}{2} (x + 1 - ln ∣ x + 1 ∣))]_{0}^{1}

Simplify

[\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - (- \frac{1}{2} (x + 1 - ln ∣ x + 1 ∣))]_{0}^{1} : [\frac{1}{2} (x^{2} ln (1 + \frac{1}{x}) + x + 1 - ln ∣ x + 1 ∣)]_{0}^{1}

= [\frac{1}{2} (x^{2} ln (1 + \frac{1}{x}) + x + 1 - ln ∣ x + 1 ∣)]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

What is the integral from 0 to 1 of xln(1+1/x) ?
The integral from 0 to 1 of xln(1+1/x) is 1/2