What is the integral from 1 to c of xln(x) ?

The integral from 1 to c of xln(x) is (c^2ln(c))/2+(-c^2+1)/4

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integral from 1 to c of xln(x)

\int_{1}^{c} x ln (x) d x

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

Solution steps

\int_{1}^{c} x ln (x) d x

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{2} r 4 - r^{2} - r^{2} d r

\int_{- 2}^{1} x^{2} - 9 d x

\int_{- 2}^{7} 1

\int_{0}^{3} x - x^{2} + 2 x d x

\int_{0}^{x} \frac{1}{x ^{2} + 1} d x

What is the integral from 1 to c of xln(x) ?
The integral from 1 to c of xln(x) is (c^2ln(c))/2+(-c^2+1)/4