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人気のある微分積分 >

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

解

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

解

\frac{1}{18} ln (7)

+1

十進法表記

0.10810 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{7}^{49} \frac{1}{18 u} d u

定数を除く:

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

= \frac{1}{18} \cdot \int_{7}^{49} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} [ln ∣ u ∣]_{7}^{49}

限度を計算する:

ln (7)

= \frac{1}{18} ln (7)

グラフ

人気の例

integral from 0 to 7 of 1/(x-2)

\int_{0}^{7} \frac{1}{x - 2} d x

integral from 3 to 10 of 1/(sqrt(x-10))

\int_{3}^{10} \frac{1}{x - 10} d x

integral from 1 to ln(2) of x*e^{4x}

\int_{1}^{l n (2)} x \cdot e^{4 x} d x

integral from 0 to r of (br+c)2pir

\int_{0}^{r} (b r + c) 2 π r d r

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

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