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受欢迎的微积分 >

integral of 1/(sqrt(12x+0.02x^2))

解答

\int \frac{1}{12 x + 0.02 x ^{2}} d x

解答

7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

求解步骤

\int \frac{1}{12 x + 0.02 x ^{2}} d x

分解

\frac{1}{12 x + 0.02 x ^{2}}

= \int \frac{1}{x ( 0.02 x + 12 )} d x

使用换元积分法

= \int \frac{1}{2 \cdot 0.1 u ( u - 12 )} d u

提出常数:

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

= \frac{1}{2 \cdot 0.1} \cdot \int \frac{1}{u ( u - 12 )} d u

化简

= 7.07106 \dots \cdot \int \frac{1}{u ( u - 12 )} d u

对

u (u - 12)

配方:

(u - 6)^{2} - 36

= 7.07106 \dots \cdot \int \frac{1}{( u - 6 ) ^{2} - 36} d u

使用换元积分法

= 7.07106 \dots \cdot \int \frac{1}{v ^{2} - 36} d v

使用三角换元法

= 7.07106 \dots \cdot \int sec (w) d w

使用积分常见定则:

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

= 7.07106 \dots ln ∣ tan (w) + sec (w) ∣

代回

= 7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6)))

化简

7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) : 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

解答补常数

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

作图

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