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人気のある微分積分 >

integral from 1 to 3 of (x^2+6)/(4x-x^2)

解

\int_{1}^{3} \frac{x ^{2} + 6}{4 x - x ^{2}} d x

解

- 2 + 7 ln (3)

+1

十進法表記

5.69028 \dots

解答ステップ

\int_{1}^{3} \frac{x ^{2} + 6}{4 x - x ^{2}} d x

因数

\frac{x ^{2} + 6}{4 x - x ^{2}}

= \int_{1}^{3} \frac{x ^{2} + 6}{- x ( x - 4 )} d x

定数を除く:

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

= - \int_{1}^{3} \frac{x ^{2} + 6}{x ( x - 4 )} d x

拡張

\frac{x ^{2} + 6}{x ( x - 4 )} : \frac{x ^{2}}{x ( x - 4 )} + \frac{6}{x ( x - 4 )}

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}^{3} \frac{x ^{2}}{x ( x - 4 )} d x + \int_{1}^{3} \frac{6}{x ( x - 4 )} d x)

\int_{1}^{3} \frac{x ^{2}}{x ( x - 4 )} d x = 2 - 4 ln (3)

\int_{1}^{3} \frac{6}{x ( x - 4 )} d x = - 3 ln (3)

= - (2 - 4 ln (3) - 3 ln (3))

簡素化

= - 2 + 7 ln (3)

グラフ

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