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人気のある微分積分 >

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

解

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

解

\frac{3 37 - 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (2 + 5)

+1

十進法表記

8.26814 \dots

解答ステップ

\int_{1}^{3} 1 + 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 (2)}^{a r c t a n (6)} \frac{1}{2} sec^{3} (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 (2)}^{a r c t a n (6)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} \cdot \int_{a r c t a n (2)}^{a r c t a n (6)} sec (u) d u)

\int_{a r c t a n (2)}^{a r c t a n (6)} sec (u) d u = ln (6 + 37) - ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

限度を計算する:

337 - 5

= \frac{1}{2} (337 - 5 + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

簡素化

= \frac{3 37 - 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (2 + 5)

グラフ

人気の例

(\partial)/(\partial x)(2x-2)

\frac{\partial}{\partial x} (2 x - 2)

limit as t approaches 0 of 3/t-3/(t^2+t)

t \to 0 lim (\frac{3}{t} - \frac{3}{t ^{2} + t})

integral of 4/(x^5)-3/(x^4)

\int \frac{4}{x ^{5}} - \frac{3}{x ^{4}} d x

integral from-1 to 1 of (3x)/((x+7)^2)

\int_{- 1}^{1} \frac{3 x}{( x + 7 ) ^{2}} d x

(\partial)/(\partial y)(cos(x/y))

\frac{\partial}{\partial y} (cos (\frac{x}{y}))