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受欢迎的微积分 >

integral of 1/(x^2sqrt(x-1))

解答

\int \frac{1}{x ^{2} x - 1} d x

解答

arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)) + C

求解步骤

\int \frac{1}{x ^{2} x - 1} d x

使用换元积分法

= \int \frac{2}{( u ^{2} + 1 ) ^{2}} d u

提出常数:

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

= 2 \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

使用三角换元法

= 2 \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

使用三角恒等式改写

= 2 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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

= 2 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 2 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= 2 \cdot \frac{1}{2} (arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)))

化简

2 \cdot \frac{1}{2} (arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))) : arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))

= arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))

解答补常数

= arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)) + C

作图

流行的例子

integral of (4^x-1/(3x))

\int (4^{x} - \frac{1}{3 x}) d x

麦克劳林 1/(1+x+x^2)

ma c l a u r in \frac{1}{1 + x + x ^{2}}

因式分解 12x^5-8x^4+20x^3

f a c t or 12 x^{5} - 8 x^{4} + 20 x^{3}

斜率 (-1,8),(3,-4)

s l o p e (- 1, 8), (3, - 4)

f(x)=(3x^3-4)(-4x-3)

f (x) = (3 x^{3} - 4) (- 4 x - 3)