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人気のある微分積分 >

integral from 0 to 4 of 1/(1+sqrt(2x+1))

解

\int_{0}^{4} \frac{1}{1 + 2 x + 1} d x

解

2 - ln (2)

+1

十進法表記

1.30685 \dots

解答ステップ

\int_{0}^{4} \frac{1}{1 + 2 x + 1} d x

u置換積分法を適用する

= \int_{1}^{3} \frac{u}{1 + u} d u

u置換積分法を適用する

= \int_{2}^{4} \frac{v - 1}{v} d v

拡張

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \int_{2}^{4} 1 - \frac{1}{v} d v

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

= \int_{2}^{4} 1 d v - \int_{2}^{4} \frac{1}{v} d v

\int_{2}^{4} 1 d v = 2

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

= 2 - ln (2)

グラフ

人気の例

integral from 2 to 8 of sqrt(x^2+1)

\int_{2}^{8} x^{2} + 1 d x

integral from 0 to pi/4 of sin^2(t)

\int_{0}^{\frac{π}{4}} sin^{2} (t) d t

integral from 1 to e of 1/(x(2x+1))

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

integral from a to b of sqrt(3-x^2)

\int_{a}^{b} 3 - x^{2} d x

integral from-pi/2 to pi/2 of (cos(x))^2

\int_{- \frac{π}{2}}^{\frac{π}{2}} (cos (x))^{2} d x