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受欢迎的微积分 >

integral of 1/((1+4x^2)(arctan(2x))^3)

解答

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

解答

- \frac{1}{4 arctan ^{2} ( 2 x )} + C

求解步骤

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

使用换元积分法

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

使用换元积分法

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

使用幂法则

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

代回

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

化简

\frac{1}{2} (- \frac{1}{2 arctan ^{2} ( 2 x )}) : - \frac{1}{4 arctan ^{2} ( 2 x )}

= - \frac{1}{4 arctan ^{2} ( 2 x )}

解答补常数

= - \frac{1}{4 arctan ^{2} ( 2 x )} + C

作图

流行的例子

integral of sin^4(6x)

\int sin^{4} (6 x) d x

limit as x approaches 0+of 3x^{sqrt(x)}

x \to 0 + lim (3 x^{x})

2x^2y^{''}+6xy^'+2y=0,y(1)=2,y^'(1)=-1

2 x^{2} y^{^{''}} + 6 x y^{^{'}} + 2 y = 0, y (1) = 2, y^{^{'}} (1) = - 1

y^{''}-6y^'+9y=2te^{3t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 t e^{3 t}

limit as x approaches 2-of 4/((x-2)^5)

x \to 2 - lim (\frac{4}{( x - 2 ) ^{5}})