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受欢迎的微积分 >

integral of sqrt(x^2+2x+5)

解答

\int x^{2} + 2 x + 5 d x

解答

4 (\frac{1}{8} (x + 1) x^{2} + 2 x + 5 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 5}{2})) + C

求解步骤

\int x^{2} + 2 x + 5 d x

对

x^{2} + 2 x + 5

配方:

(x + 1)^{2} + 4

= \int (x + 1)^{2} + 4 d x

使用换元积分法

= \int u^{2} + 4 d u

使用三角换元法

= \int 4 sec^{3} (v) d v

提出常数:

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

= 4 \cdot \int sec^{3} (v) d v

使用降次积分法

= 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

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

代回

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

化简

4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x + 1 ) ) ) sin ( arctan ( \frac{1}{2} ( x + 1 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1)))) : 4 (\frac{1}{8} (x + 1) x^{2} + 2 x + 5 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 5}{2}))

= 4 (\frac{1}{8} (x + 1) x^{2} + 2 x + 5 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 5}{2}))

解答补常数

= 4 (\frac{1}{8} (x + 1) x^{2} + 2 x + 5 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 5}{2})) + C

作图

流行的例子

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t an g e n t f (x) = x^{2} + 8, (4, 24)

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\frac{d}{d x} (\frac{3 - 2 x}{3 + 2 x})

integral of x^4sin(3x)

\int x^{4} sin (3 x) d x

integral of-(x^2+2)e^x

\int - (x^{2} + 2) e^{x} d x

(\partial)/(\partial x)(3^{2y})

\frac{\partial}{\partial x} (3^{2 y})