What is the integral of xln(x^{-1}) ?

The integral of xln(x^{-1}) is -1/2 x^2ln(x)+(x^2)/4+C

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integral of xln(x^{-1})

\int x ln (x^{- 1}) d x

- \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4} + C

Solution steps

\int x ln (x^{- 1}) d x

Simplify

x ln (x^{- 1}) : - x ln (x)

= \int - x ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int x ln (x) d x

Apply Integration By Parts

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

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

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

Simplify

= - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}

Add a constant to the solution

= - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4} + C

What is the integral of xln(x^{-1}) ?
The integral of xln(x^{-1}) is -1/2 x^2ln(x)+(x^2)/4+C