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受欢迎的微积分 >

integral of sqrt(4x^2-4x+10)

解答

\int 4 x^{2} - 4 x + 10 d x

解答

\frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3})) + C

求解步骤

\int 4 x^{2} - 4 x + 10 d x

对

4 x^{2} - 4 x + 10

配方:

4 (x - \frac{1}{2})^{2} + 9

= \int 4 (x - \frac{1}{2})^{2} + 9 d x

使用换元积分法

= \int 4 u^{2} + 9 d u

使用三角换元法

= \int \frac{9}{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{9}{2} \cdot \int sec^{3} (v) d v

使用降次积分法

= \frac{9}{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{9}{2} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

代回

= \frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2}))))

化简

\frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2})))) : \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3}))

= \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3}))

解答补常数

= \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3})) + C

作图

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