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人気のある微分積分 >

integral from 0 to x of ln(1+x)

解

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

解

x ln (x + 1) - x + ln (x + 1)

解答ステップ

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

u置換積分法を適用する

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

部分積分を適用する

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

\int 1 d u = u

= [u ln (u) - u]_{1}^{1 + x}

限度を計算する:

x ln (x + 1) - x + ln (x + 1)

= x ln (x + 1) - x + ln (x + 1)

人気の例

integral from 0 to 0.3 of xln(1+x^3)

\int_{0}^{0.3} x ln (1 + x^{3}) d x

integral from-2 to 2 of x^5

\int_{- 2}^{2} x^{5} d x

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

\int_{2}^{x^{3}} \frac{1}{t ^{3}} d t

integral from 0 to 5 of 6

\int_{0}^{5} 6

integral from 1 to 2 of (4x^3-6)

\int_{1}^{2} (4 x^{3} - 6) d x