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人気のある微分積分 >

integral of sqrt(1+100x^2)

解

\int 1 + 100 x^{2} d x

解

\frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2}) + C

解答ステップ

\int 1 + 100 x^{2} d x

三角関数による置換を適用する

= \int \frac{1}{10} 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}{10} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{1}{10} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{10} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (10 x)

= \frac{1}{10} (\frac{sec ^{2} ( arctan ( 10 x ) ) sin ( arctan ( 10 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (10 x)) + sec (arctan (10 x)) ∣)

簡素化

\frac{1}{10} (\frac{sec ^{2} ( arctan ( 10 x ) ) sin ( arctan ( 10 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (10 x)) + sec (arctan (10 x)) ∣) : \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2})

= \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2})

定数を解答に追加する

= \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2}) + C

グラフ

人気の例

\frac{1}{2} y^{^{'}} + 2 y = 0

integral from x to infinity of 2e^{-x-y}

\int_{x}^{\infty} 2 e^{- x - y} d y

derivative y=4x^2-13x-9/(x^4)

d er i v a t i v e y = 4 x^{2} - 13 x - \frac{9}{x ^{4}}

tangent f(x)= 2/x ,(-3,-2/3)

t an g e n t f (x) = \frac{2}{x}, (- 3, - \frac{2}{3})

derivative of x-ln(x^2)

\frac{d}{d x} (x - ln (x^{2}))