zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 2/(x^3sqrt(x^4-25))

解答

\int \frac{2}{x ^{3} x ^{4} - 25} d x

解答

\frac{x ^{4} - 25}{25 x ^{2}} + C

求解步骤

\int \frac{2}{x ^{3} x ^{4} - 25} d x

提出常数:

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

= 2 \cdot \int \frac{1}{x ^{3} x ^{4} - 25} d x

使用换元积分法

= 2 \cdot \int \frac{1}{2 u ^{2} u ^{2} - 25} d u

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 \frac{1}{u ^{2} u ^{2} - 25} d u

使用三角换元法

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{25 sec ( v )} 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 \frac{1}{25} \cdot \int \frac{1}{sec ( v )} d v

使用三角恒等式改写

= 2 \cdot \frac{1}{2} \cdot \frac{1}{25} \cdot \int cos (v) d v

使用积分常见定则:

\int cos (v) d v = sin (v)

= 2 \cdot \frac{1}{2} \cdot \frac{1}{25} sin (v)

代回

= 2 \cdot \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} x^{2}))

化简

2 \cdot \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} x^{2})) : \frac{x ^{4} - 25}{25 x ^{2}}

= \frac{x ^{4} - 25}{25 x ^{2}}

解答补常数

= \frac{x ^{4} - 25}{25 x ^{2}} + C

作图

流行的例子

integral of 1/(x^3sqrt(x^2-4))

\int \frac{1}{x ^{3} x ^{2} - 4} d x

integral of 9x^2-30x+15

\int 9 x^{2} - 30 x + 15 d x

(\partial)/(\partial x)(x*sin(y-z))

\frac{\partial}{\partial x} (x \cdot sin (y - z))

d/(dt)((\sqrt[3]{t})/5)

\frac{d}{d t} (\frac{3 t}{5})

(\partial)/(\partial x)(6-(x^2+y^2)^{1/5})

\frac{\partial}{\partial x} (6 - (x^{2} + y^{2})^{\frac{1}{5}})