integral from 1 to n of xln(x)

解

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

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

解答ステップ

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

部分積分を適用する

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

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

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

限度を計算する:

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

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

簡素化

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