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人気のある微分積分 >

integral from 1 to 2 of 5/(x^3+2x)

解

\int_{1}^{2} \frac{5}{x ^{3} + 2 x} d x

解

\frac{5}{4} ln (2)

+1

十進法表記

0.86643 \dots

解答ステップ

\int_{1}^{2} \frac{5}{x ^{3} + 2 x} d x

定数を除く:

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

= 5 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 2 x} d x

以下の部分分数を得る:

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

= 5 \cdot \int_{1}^{2} \frac{1}{2 x} - \frac{x}{2 ( x ^{2} + 2 )} 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

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

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

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

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

簡素化

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

= \frac{5}{4} ln (2)

グラフ

人気の例

integral from 0 to 1 of (5^x-3^x)

\int_{0}^{1} (5^{x} - 3^{x}) d x

tangent y=2x^3,(1,2)

t an g e n t y = 2 x^{3}, (1, 2)

derivative f(x)=sqrt(6x+3)

d er i v a t i v e f (x) = 6 x + 3

integral of (ln(11x))/(x^9)

\int \frac{ln ( 11 x )}{x ^{9}} d x

integral of ((7x^2+5x-6))/((x^2-4)(x+2))

\int \frac{( 7 x ^{2} + 5 x - 6 )}{( x ^{2} - 4 ) ( x + 2 )} d x