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

解答

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

\frac{3 x ^{4} arctan ( \frac{x}{a} ) + 3 a x ^{3} + 6 a ^{2} x ^{2} arctan ( \frac{x}{a} ) + 5 a ^{3} x + 3 a ^{4} arctan ( \frac{x}{a} )}{8 a ^{5} ( x ^{2} + a ^{2} ) ^{2}} + C

求解步骤

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

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

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

使用换元积分法

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

提出常数:

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

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

使用降次积分法

= \frac{1}{a ^{5}} (\frac{3}{4} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

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

= \frac{1}{a ^{5}} (\frac{3}{4} \cdot \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

u = \frac{x}{a}

代回

= \frac{1}{a ^{5}} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{a}) + \frac{\frac{x}{a}}{1 + ( \frac{x}{a} ) ^{2}}) + \frac{\frac{x}{a}}{2 \cdot 2 ( ( \frac{x}{a} ) ^{2} + 1 ) ^{2}}

化简

\frac{1}{a ^{5}} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{a}) + \frac{\frac{x}{a}}{1 + ( \frac{x}{a} ) ^{2}}) + \frac{\frac{x}{a}}{2 \cdot 2 ( ( \frac{x}{a} ) ^{2} + 1 ) ^{2}} : \frac{3 x ^{4} arctan ( \frac{x}{a} ) + 3 a x ^{3} + 6 a ^{2} x ^{2} arctan ( \frac{x}{a} ) + 5 a ^{3} x + 3 a ^{4} arctan ( \frac{x}{a} )}{8 a ^{5} ( x ^{2} + a ^{2} ) ^{2}}

= \frac{3 x ^{4} arctan ( \frac{x}{a} ) + 3 a x ^{3} + 6 a ^{2} x ^{2} arctan ( \frac{x}{a} ) + 5 a ^{3} x + 3 a ^{4} arctan ( \frac{x}{a} )}{8 a ^{5} ( x ^{2} + a ^{2} ) ^{2}}

解答补常数

= \frac{3 x ^{4} arctan ( \frac{x}{a} ) + 3 a x ^{3} + 6 a ^{2} x ^{2} arctan ( \frac{x}{a} ) + 5 a ^{3} x + 3 a ^{4} arctan ( \frac{x}{a} )}{8 a ^{5} ( x ^{2} + a ^{2} ) ^{2}} + C

\int \frac{2 x + 1}{x ^{3} + 9 x} d x

\int 3 a x^{2} d x

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

\int 9 x - 3 d x

\int \frac{3 x - 1}{x + 2} d x