integral from 1 to 10 of 1/(xIn^2x)

解

\int_{1}^{10} \frac{1}{x I n ^{2} x} d x

\frac{9}{10 I n ^{2}}

解答ステップ

\int_{1}^{10} \frac{1}{x I n ^{2} x} d x

定数を除く:

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

= \frac{1}{I n ^{2}} \cdot \int_{1}^{10} \frac{1}{xx} d x

xx = x^{2}

= \frac{1}{I n ^{2}} \cdot \int_{1}^{10} \frac{1}{x ^{2}} d x

べき乗則を適用する

= \frac{1}{I n ^{2}} [- \frac{1}{x}]_{1}^{10}

簡素化

\frac{1}{I n ^{2}} [- \frac{1}{x}]_{1}^{10} : \frac{[ - \frac{1}{x} ] _{1}^{10}}{I n ^{2}}

= \frac{[ - \frac{1}{x} ] _{1}^{10}}{I n ^{2}}

限度を計算する:

\frac{9}{10}

= \frac{\frac{9}{10}}{I n ^{2}}

簡素化

= \frac{9}{10 I n ^{2}}