integral of (1+x^2)*sqrt(1+64x^2)

解答

\int (1 + x^{2}) \cdot 1 + 64 x^{2} d x

\frac{255}{512} x 64 x^{2} + 1 + \frac{1}{256} x (64 x^{2} + 1)^{\frac{3}{2}} + \frac{255}{4096} ln 8 x + 64 x^{2} + 1 + C

求解步骤

\int (1 + x^{2}) 1 + 64 x^{2} d x

乘开

(1 + x^{2}) 1 + 64 x^{2} : 1 + 64 x^{2} + x^{2} 1 + 64 x^{2}

= \int 1 + 64 x^{2} + x^{2} 1 + 64 x^{2} d x

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

= \int 1 + 64 x^{2} d x + \int x^{2} 1 + 64 x^{2} d x

\int 1 + 64 x^{2} d x = \frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2})

\int x^{2} 1 + 64 x^{2} d x = \frac{1}{4096} (- 8 x 1 + 64 x^{2} - ln 8 x + 1 + 64 x^{2} + 16 x (64 x^{2} + 1)^{\frac{3}{2}})

= \frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2}) + \frac{1}{4096} (- 8 x 1 + 64 x^{2} - ln 8 x + 1 + 64 x^{2} + 16 x (64 x^{2} + 1)^{\frac{3}{2}})

化简

\frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2}) + \frac{1}{4096} (- 8 x 1 + 64 x^{2} - ln 8 x + 1 + 64 x^{2} + 16 x (64 x^{2} + 1)^{\frac{3}{2}}) : \frac{255}{512} x 64 x^{2} + 1 + \frac{1}{256} x (64 x^{2} + 1)^{\frac{3}{2}} + \frac{255}{4096} ln 8 x + 64 x^{2} + 1

= \frac{255}{512} x 64 x^{2} + 1 + \frac{1}{256} x (64 x^{2} + 1)^{\frac{3}{2}} + \frac{255}{4096} ln 8 x + 64 x^{2} + 1

解答补常数

= \frac{255}{512} x 64 x^{2} + 1 + \frac{1}{256} x (64 x^{2} + 1)^{\frac{3}{2}} + \frac{255}{4096} ln 8 x + 64 x^{2} + 1 + C

(2 y^{2} x + 3 x) d x + (2 x^{2} y) d y = 0

\frac{\partial}{\partial v} (\frac{v ^{2}}{u})

\frac{d}{d x} (2 - x - y + 1)

x \to 1 lim (1 + x^{2})

\frac{d}{d x} (100)