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人気のある微分積分 >

integral from 0 to 1 of 1/(sqrt(x+x^2))

解

\int_{0}^{1} \frac{1}{x + x ^{2}} d x

解

2 ln (1 + 2)

+1

十進法表記

1.76274 \dots

解答ステップ

\int_{0}^{1} \frac{1}{x + x ^{2}} d x

因数

\frac{1}{x + x ^{2}}

= \int_{0}^{1} \frac{1}{x ( x + 1 )} d x

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (x + 1) = x (x + 1)

= \int_{0}^{1} \frac{1}{x x + 1} d x

u置換積分法を適用する

= \int_{0}^{1} \frac{2}{u ^{2} + 1} d u

定数を除く:

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

= 2 \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

共通積分を使用する:

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= 2 [ln u^{2} + 1 + u]_{0}^{1}

限度を計算する:

ln (1 + 2)

= 2 ln (1 + 2)

グラフ

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