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受欢迎的微积分 >

integral of sqrt(1+(2x-4)^2)

解答

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

解答

\frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

求解步骤

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

乘开

1 + (2 x - 4)^{2} : 4 x^{2} - 16 x + 17

= \int 4 x^{2} - 16 x + 17 d x

对

4 x^{2} - 16 x + 17

配方:

4 (x - 2)^{2} + 1

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

使用换元积分法

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

使用三角换元法

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

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} (v) d v

使用降次积分法

= \frac{1}{2} (\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) ∣

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

代回

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

化简

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 ( x - 2 ) ) ) sin ( arctan ( 2 ( x - 2 ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 (x - 2))) + sec (arctan (2 (x - 2))) ∣) : \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

解答补常数

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

作图

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