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人気のある微分積分 >

integral from 4 to x of 1/(1+t^2)

解

\int_{4}^{x} \frac{1}{1 + t ^{2}} d t

解

arctan (x) - arctan (4)

解答ステップ

\int_{4}^{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)]_{4}^{x}

限度を計算する:

arctan (x) - arctan (4)

= arctan (x) - arctan (4)

人気の例

integral from 0 to 1 of x^4-x^2

\int_{0}^{1} x^{4} - x^{2} d x

integral from 0 to pi/4 of 1 (cos^4(x))

\int_{0}^{\frac{π}{4}} \frac{d}{1} (cos^{4} (x)) d x

integral from 2 to 6 of (x^2+7x-5)(x-2)

\int_{2}^{6} (x^{2} + 7 x - 5) (x - 2) d x

integral from 1 to 2 of (x-1)(2-x)

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

integral from-pi/4 to pi/4 of sin^2(x)

\int_{- \frac{π}{4}}^{\frac{π}{4}} sin^{2} (x) d x