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人気のある微分積分 >

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

解

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

解

arctan (x) - \frac{π}{4}

解答ステップ

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

共通積分を使用する:

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

= [arctan (t)]_{1}^{x}

限度を計算する:

arctan (x) - \frac{π}{4}

= arctan (x) - \frac{π}{4}

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