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)/(sqrt((1+4x)^5))

解答

\int \frac{x ^{2}}{( 1 + 4 x ) ^{5}} d x

解答

\frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}} + C

求解步骤

\int \frac{x ^{2}}{( 1 + 4 x ) ^{5}} d x

(1 + 4 x)^{5} = (1 + 4 x)^{\frac{5}{2}},

假定

(1 + 4 x) \geq 0

= \int \frac{x ^{2}}{( 1 + 4 x ) ^{\frac{5}{2}}} d x

使用换元积分法

= \int \frac{( u - 1 ) ^{2}}{64 u ^{\frac{5}{2}}} d u

提出常数:

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

= \frac{1}{64} \cdot \int \frac{( u - 1 ) ^{2}}{u ^{\frac{5}{2}}} d u

乘开

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

= \frac{1}{64} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{2}{u ^{\frac{3}{2}}} + \frac{1}{u ^{\frac{5}{2}}} 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}{64} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{2}{u ^{\frac{3}{2}}} d u + \int \frac{1}{u ^{\frac{5}{2}}} d u)

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

\int \frac{2}{u ^{\frac{3}{2}}} d u = - \frac{4}{u}

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

= \frac{1}{64} (2 u^{\frac{1}{2}} - (- \frac{4}{u}) - \frac{2}{3 u ^{\frac{3}{2}}})

u = 1 + 4 x

代回

= \frac{1}{64} (2 (1 + 4 x)^{\frac{1}{2}} - (- \frac{4}{1 + 4 x}) - \frac{2}{3 ( 1 + 4 x ) ^{\frac{3}{2}}})

化简

\frac{1}{64} (2 (1 + 4 x)^{\frac{1}{2}} - (- \frac{4}{1 + 4 x}) - \frac{2}{3 ( 1 + 4 x ) ^{\frac{3}{2}}}) : \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}}

= \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}}

解答补常数

= \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}} + C

作图

流行的例子

integral of x/(1+sqrt(1+x))

\int \frac{x}{1 + 1 + x} d x

integral of (7x^3-6x+4)/(x^3-x^2)

\int \frac{7 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

integral of (ln^2(x+1))/(x+1)

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

integral of x^3(2x+1/x)

\int x^{3} (2 x + \frac{1}{x}) d x

integral of (sin(3x))/(1+cos(3x))

\int \frac{sin ( 3 x )}{1 + cos ( 3 x )} d x