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 sqrt(1+x^2)(x^5)

解答

\int 1 + x^{2} (x^{5}) d x

解答

\frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

求解步骤

\int 1 + x^{2} x^{5} d x

使用换元积分法

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

提出常数:

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

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

使用换元积分法

= \frac{1}{2} \cdot \int (v - 1)^{2} v d v

乘开

(v - 1)^{2} v : v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + v

= \frac{1}{2} \cdot \int v^{\frac{5}{2}} - 2 v^{\frac{3}{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

= \frac{1}{2} (\int v^{\frac{5}{2}} d v - \int 2 v^{\frac{3}{2}} d v + \int v d v)

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

\int 2 v^{\frac{3}{2}} d v = \frac{4}{5} v^{\frac{5}{2}}

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{7} v^{\frac{7}{2}} - \frac{4}{5} v^{\frac{5}{2}} + \frac{2}{3} v^{\frac{3}{2}})

代回

= \frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}})

化简

\frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

解答补常数

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

作图

流行的例子

integral of sin^2((3x)/4)

\int sin^{2} (\frac{3 x}{4}) d x

integral of cos(5x)sin(2x)

\int cos (5 x) sin (2 x) d x

integral of cos((npix)/7)

\int cos (\frac{nπ x}{7}) d x

integral of 7/(1+16x^2)

\int \frac{7}{1 + 16 x ^{2}} d x

integral of-e^{2x}cos(x)

\int - e^{2 x} cos (x) d x