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受欢迎的微积分 >

integral of sqrt(4r^2+1)

解答

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

解答

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

求解步骤

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

使用三角换元法

= \int \frac{1}{2} sec^{3} (u) d u

提出常数:

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

= \frac{1}{2} \cdot \int sec^{3} (u) d u

使用降次积分法

= \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) ∣)

u = arctan (2 r)

代回

= \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)) ∣)

化简

\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})

解答补常数

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

作图

流行的例子

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

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泰勒 sqrt(1+2x)

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

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