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人気のある微分積分 >

integral of sqrt(9+16t^2)

解

\int 9 + 16 t^{2} d t

解

\frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3})) + C

解答ステップ

\int 9 + 16 t^{2} d t

三角関数による置換を適用する

= \int \frac{9}{4} 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{9}{4} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{9}{4} (\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{9}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{4}{3} t)

= \frac{9}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{3} t ) ) sin ( arctan ( \frac{4}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{3} t)) + sec (arctan (\frac{4}{3} t)))

簡素化

\frac{9}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{3} t ) ) sin ( arctan ( \frac{4}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{3} t)) + sec (arctan (\frac{4}{3} t))) : \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3}))

= \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3}))

定数を解答に追加する

= \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3})) + C

グラフ

人気の例

limit as x approaches 3 of (1-x)/(x-3)

x \to 3 lim (\frac{1 - x}{x - 3})

(\partial)/(\partial y)(x^2+2y)

\frac{\partial}{\partial y} (x^{2} + 2 y)

derivative of x^n(ln(x)^2)

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

derivative f(x)=(9-xe^x)/(x+e^x)

d er i v a t i v e f (x) = \frac{9 - x e ^{x}}{x + e ^{x}}

derivative of (5x-2^3)

\frac{d}{d x} ((5 x - 2)^{3})