integral from 1 to 2 of (4x^2-3x+4)/x

解

\int_{1}^{2} \frac{4 x ^{2} - 3 x + 4}{x} d x

4 ln (2) + 3

十進法表記

5.77258 \dots

解答ステップ

\int_{1}^{2} \frac{4 x ^{2} - 3 x + 4}{x} d x

拡張

\frac{4 x ^{2} - 3 x + 4}{x} : 4 x - 3 + \frac{4}{x}

= \int_{1}^{2} 4 x - 3 + \frac{4}{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} 4 x d x - \int_{1}^{2} 3 d x + \int_{1}^{2} \frac{4}{x} d x

\int_{1}^{2} 4 x d x = 6

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

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

= 6 - 3 + 4 ln (2)

簡素化

= 4 ln (2) + 3