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人気のある微分積分 >

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

解

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

解

arctan (t)

解答ステップ

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

共通積分を使用する:

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

= [arctan (x)]_{0}^{t}

限度を計算する:

arctan (t)

= arctan (t)

人気の例

integral from 0 to 1 of 4xarctan(x)

\int_{0}^{1} 4 x arctan (x) d x

integral from 1 to 4 of e^{-x}+(ln(x))/x

\int_{1}^{4} e^{- x} + \frac{ln ( x )}{x} d x

integral from 0 to 1 of log_{e}(x)

\int_{0}^{1} lo g_{e} (x) d x

integral from 3 to 5 of 1/(x^2-4x+8)

\int_{3}^{5} \frac{1}{x ^{2} - 4 x + 8} d x

integral from 0 to 2 of x^2 3/4 x(2-x)

\int_{0}^{2} x^{2} \frac{3}{4} x (2 - x) d x