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人気のある微分積分 >

integral from 1 to 4 of 1/(x(2+sqrt(x)))

解

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

解

- ln (2) + ln (3)

+1

十進法表記

0.40546 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{2} \frac{2}{u ( u + 2 )} d u

定数を除く:

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

= 2 \cdot \int_{1}^{2} \frac{1}{u ( u + 2 )} d u

以下の部分分数を得る:

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= 2 \cdot \int_{1}^{2} \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} d u

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

= 2 (\int_{1}^{2} \frac{1}{2 u} d u - \int_{1}^{2} \frac{1}{2 ( u + 2 )} d u)

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

\int_{1}^{2} \frac{1}{2 ( u + 2 )} d u = ln (2) - \frac{1}{2} ln (3)

= 2 (\frac{1}{2} ln (2) - (ln (2) - \frac{1}{2} ln (3)))

簡素化

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

= - ln (2) + ln (3)

グラフ

人気の例

integral from 3 to 4 of (e^{ln(x)})/x

\int_{3}^{4} \frac{e ^{l n (x)}}{x} d x

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

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

integral from 2 to 4 of 2x^3+2x

\int_{2}^{4} 2 x^{3} + 2 x d x

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

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

integral from-64 to 8 of 1/(\sqrt[3]{x)}

\int_{- 64}^{8} \frac{1}{3 x} d x