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 (x^2)/(9+4x^2)

解答

\int \frac{x ^{2}}{9 + 4 x ^{2}} d x

解答

\frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x)) + C

求解步骤

\int \frac{x ^{2}}{9 + 4 x ^{2}} d x

9 + 4 x^{2} = 9 (1 + \frac{4 x ^{2}}{9})

= \int \frac{x ^{2}}{9 ( 1 + \frac{4 x ^{2}}{9} )} d x

提出常数:

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

= \frac{1}{9} \cdot \int \frac{x ^{2}}{1 + \frac{4 x ^{2}}{9}} d x

使用换元积分法

= \frac{1}{9} \cdot \int \frac{27 u ^{2}}{8 ( 1 + u ^{2} )} d u

提出常数:

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

= \frac{1}{9} \cdot \frac{27}{8} \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= \frac{1}{9} \cdot \frac{27}{8} \cdot \int - \frac{1}{1 + u ^{2}} + 1 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

= \frac{1}{9} \cdot \frac{27}{8} (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

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

\int 1 d u = u

= \frac{1}{9} \cdot \frac{27}{8} (- arctan (u) + u)

u = \frac{2}{3} x

代回

= \frac{1}{9} \cdot \frac{27}{8} (- arctan (\frac{2}{3} x) + \frac{2}{3} x)

化简

\frac{1}{9} \cdot \frac{27}{8} (- arctan (\frac{2}{3} x) + \frac{2}{3} x) : \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x))

= \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x))

解答补常数

= \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x)) + C

作图

流行的例子

integral of 3x^2sin(x^3-1)

\int 3 x^{2} sin (x^{3} - 1) d x

derivative-4t(3t^5-1)^4

d er i v a t i v e - 4 t (3 t^{5} - 1)^{4}

derivative of ln(xsqrt(x^2-10))

\frac{d}{d x} (ln (x x^{2} - 10))

tangent f(x)=4x^2-3x+2

t an g e n t f (x) = 4 x^{2} - 3 x + 2

integral from 0 to 5 of x^2

\int_{0}^{5} x^{2} d x