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integral of 10sqrt(5x^2-1)

解答

\int 10 5 x^{2} - 1 d x

解答

5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1 + C

求解步骤

\int 10 5 x^{2} - 1 d x

提出常数:

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

= 10 \cdot \int 5 x^{2} - 1 d x

使用三角换元法

= 10 \cdot \int \frac{1}{5} tan^{2} (u) sec (u) d u

提出常数:

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

= 10 \cdot \frac{1}{5} \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= 10 \cdot \frac{1}{5} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 10 \cdot \frac{1}{5} \cdot \int - sec (u) + sec^{3} (u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 10 \cdot \frac{1}{5} (- \int sec (u) d u + \int sec^{3} (u) d u)

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

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 10 \cdot \frac{1}{5} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arcsec (5 x)

代回

= 10 \cdot \frac{1}{5} (- ln tan (arcsec (5 x)) + sec (arcsec (5 x)) + \frac{1}{2} sec (arcsec (5 x)) tan (arcsec (5 x)) + \frac{1}{2} ln tan (arcsec (5 x)) + sec (arcsec (5 x)))

化简

10 \cdot \frac{1}{5} (- ln tan (arcsec (5 x)) + sec (arcsec (5 x)) + \frac{1}{2} sec (arcsec (5 x)) tan (arcsec (5 x)) + \frac{1}{2} ln tan (arcsec (5 x)) + sec (arcsec (5 x))) : 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1

= 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1

解答补常数

= 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1 + C

作图

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