integral from 1 to 2 of ((3x^2+9x+1)/x)

解

\int_{1}^{2} (\frac{3 x ^{2} + 9 x + 1}{x}) d x

ln (2) + \frac{27}{2}

十進法表記

14.19314 \dots

解答ステップ

\int_{1}^{2} \frac{3 x ^{2} + 9 x + 1}{x} d x

拡張

\frac{3 x ^{2} + 9 x + 1}{x} : 3 x + 9 + \frac{1}{x}

= \int_{1}^{2} 3 x + 9 + \frac{1}{x} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 3 x d x + \int_{1}^{2} 9 d x + \int_{1}^{2} \frac{1}{x} d x

\int_{1}^{2} 3 x d x = \frac{9}{2}

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

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

= \frac{9}{2} + 9 + ln (2)

分数を組み合わせる

9 + \frac{9}{2} : \frac{27}{2}

= ln (2) + \frac{27}{2}