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人気のある微分積分 >

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

解

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

解

- 2 1 + e + ln (\frac{2 + 2 e + 1 + e}{2 + e - 2 e + 1}) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

+1

十進法表記

0.57245 \dots

解答ステップ

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

u置換積分法を適用する

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

部分積分を適用する

= [2 ln (u) 1 + u - \int \frac{2 1 + u}{u} d u]_{1}^{e}

\int \frac{2 1 + u}{u} d u = - 4 (\frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1 - 1 + u)

= [2 ln (u) 1 + u - (- 4 (\frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1 - 1 + u))]_{1}^{e}

簡素化

= [2 ln (u) 1 + u + 4 (\frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1 - 1 + u)]_{1}^{e}

限度を計算する:

- 2 1 + e + 2 ln (e + 1 + 1) - 2 ln (e + 1 - 1) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

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

簡素化

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

グラフ

人気の例

integral from 0 to pi/3 of tan(θ)

\int_{0}^{\frac{π}{3}} tan (θ) d θ

integral from-pi to pi of (x^2+x)sin(x)

\int_{- π}^{π} (x^{2} + x) sin (x) d x

integral from-1 to 0 of 5(e^{1/x})/(x^3)

\int_{- 1}^{0} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

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

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

integral from-5 to-1 of 1/16

\int_{- 5}^{- 1} \frac{1}{16} d x