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人気のある微分積分 >

integral from 1 to 3 of 1/(xsqrt(4+x^2))

解

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

解

\frac{1}{2} (ln (tan (\frac{1}{2} arctan (\frac{3}{2}))) - ln (tan (\frac{1}{2} arctan (\frac{1}{2}))))

+1

十進法表記

0.40924 \dots

解答ステップ

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

三角関数による置換を適用する

= \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{3}{2})} \frac{sec ( u )}{2 tan ( u )} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{3}{2})} \frac{sec ( u )}{tan ( u )} d u

u置換積分法を適用する

= \frac{1}{2} \cdot \int_{t a n (\frac{a r c t a n ( \frac{1}{2} )}{2})}^{t a n (\frac{a r c t a n ( \frac{3}{2} )}{2})} \frac{1}{v} d v

共通積分を使用する:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} [ln ∣ v ∣]_{t a n (\frac{a r c t a n ( \frac{1}{2} )}{2})}^{t a n (\frac{a r c t a n ( \frac{3}{2} )}{2})}

限度を計算する:

ln (tan (\frac{1}{2} arctan (\frac{3}{2}))) - ln (tan (\frac{1}{2} arctan (\frac{1}{2})))

= \frac{1}{2} (ln (tan (\frac{1}{2} arctan (\frac{3}{2}))) - ln (tan (\frac{1}{2} arctan (\frac{1}{2}))))

グラフ

人気の例

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

\int_{- 2}^{- 1} (2 x^{3} + \frac{1}{x ^{2}}) d x

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

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

integral from 1 to 4 of 5x^4

\int_{1}^{4} 5 x^{4} d x

integral from 0 to 1 of ((x+5)/2)^2

\int_{0}^{1} (\frac{x + 5}{2})^{2} d x

integral from 1 to 3 of x^3+x+6

\int_{1}^{3} x^{3} + x + 6 d x