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人気のある微分積分 >

integral from 1 to 2 of (15)/(x^3+4x)

解

\int_{1}^{2} \frac{15}{x ^{3} + 4 x} d x

解

15 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

+1

十進法表記

1.71804 \dots

解答ステップ

\int_{1}^{2} \frac{15}{x ^{3} + 4 x} d x

定数を除く:

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

= 15 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

以下の部分分数を得る:

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= 15 \cdot \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} 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

= 15 (\int_{1}^{2} \frac{1}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x)

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

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= 15 (\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)))

結合

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= 15 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

グラフ

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