What is the integral from 0 to 1 of (x-1)ln(x) ?

The integral from 0 to 1 of (x-1)ln(x) is 3/4

Popular >

integral from 0 to 1 of (x-1)ln(x)

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

\frac{3}{4}

Decimal Notation

0.75

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{2} (x - 2) d x = \frac{x ^{2}}{4} - x

= [ln (x) (\frac{x ^{2}}{2} - x) - (\frac{x ^{2}}{4} - x)]_{0}^{1}

Simplify

[ln (x) (\frac{x ^{2}}{2} - x) - (\frac{x ^{2}}{4} - x)]_{0}^{1} : [\frac{1}{2} x^{2} ln (x) - x ln (x) - \frac{x ^{2}}{4} + x]_{0}^{1}

= [\frac{1}{2} x^{2} ln (x) - x ln (x) - \frac{x ^{2}}{4} + x]_{0}^{1}

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

\int_{0}^{\frac{3 π}{2}} x sin (x) d x

\int_{1}^{7} ∣ sin (x) ∣ d x

\int_{0}^{1} (1 - x) sin (nπ x) d x

\int_{\frac{π}{2}}^{\frac{3 π}{2}} x sin (x) d x

\int_{2}^{3} \frac{x}{1 - x ^{2}} d x

What is the integral from 0 to 1 of (x-1)ln(x) ?
The integral from 0 to 1 of (x-1)ln(x) is 3/4