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受欢迎的微积分 >

integral of e^xsqrt(e^{2x)+2e^x+5}

解答

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

解答

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

求解步骤

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

使用换元积分法

= \int u^{2} + 2 u + 5 d u

对

u^{2} + 2 u + 5

配方:

(u + 1)^{2} + 4

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

使用换元积分法

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

使用三角换元法

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

提出常数:

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

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

使用降次积分法

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

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

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

代回

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

化简

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

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

解答补常数

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

作图

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