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人気のある微分積分 >

integral from 0 to x of ln(z+2)

解

\int_{0}^{x} ln (z + 2) d z

解

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

解答ステップ

\int_{0}^{x} ln (z + 2) d z

u置換積分法を適用する

= \int_{2}^{x + 2} ln (u) d u

部分積分を適用する

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

\int 1 d u = u

= [u ln (u) - u]_{2}^{x + 2}

限度を計算する:

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

= x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

人気の例

integral from 0 to x of t

\int_{0}^{x} t d t

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

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

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integral from 0 to 1 of 8cos((pit)/2)

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integral from 0 to 1 of 6x^2

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