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人気のある微分積分 >

integral of sqrt(4r^2+1)

解

\int 4 r^{2} + 1 d r

解

\frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2}) + C

解答ステップ

\int 4 r^{2} + 1 d r

三角関数による置換を適用する

= \int \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 sec^{3} (u) d u

積分換算を適用する

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

代用を戻す

u = arctan (2 r)

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

簡素化

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 r ) ) sin ( arctan ( 2 r ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 r)) + sec (arctan (2 r)) ∣) : \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2})

= \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2})

定数を解答に追加する

= \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2}) + C

グラフ

人気の例

integral from-1/2 to 0 of (1/2+2y+3z)

\int_{- \frac{1}{2}}^{0} (\frac{1}{2} + 2 y + 3 z) d y

(\partial)/(\partial x)(e^{xsin(y)})

\frac{\partial}{\partial x} (e^{x s i n (y)})

テイラー sqrt(1+2x)

t a y l or 1 + 2 x

limit as x approaches 2 of (3x+4)/(8x-1)

x \to 2 lim (\frac{3 x + 4}{8 x - 1})

integral of 1/(usqrt(3-u^2))

\int \frac{1}{u 3 - u ^{2}} d u