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

解答

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

- \frac{1}{2} sin (2 arctan (x + 1)) + \frac{3 ( arctan ( x + 1 ) ( x ^{2} + 2 x + 2 ) + x + 1 )}{8 ( x ^{2} + 2 x + 2 )} + \frac{x + 1}{4 ( x ^{2} + 2 x + 2 ) ^{2}} + C

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

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

\int \frac{1}{( x ^{2} + 2 x + 2 ) ^{3}} d x = \frac{3 ( arctan ( x + 1 ) ( x ^{2} + 2 x + 2 ) + x + 1 )}{8 ( x ^{2} + 2 x + 2 )} + \frac{x + 1}{4 ( x ^{2} + 2 x + 2 ) ^{2}}

= arctan (x + 1) - (arctan (x + 1) + \frac{1}{2} sin (2 arctan (x + 1))) + \frac{3 ( arctan ( x + 1 ) ( x ^{2} + 2 x + 2 ) + x + 1 )}{8 ( x ^{2} + 2 x + 2 )} + \frac{x + 1}{4 ( x ^{2} + 2 x + 2 ) ^{2}}

化简

= - \frac{1}{2} sin (2 arctan (x + 1)) + \frac{3 ( arctan ( x + 1 ) ( x ^{2} + 2 x + 2 ) + x + 1 )}{8 ( x ^{2} + 2 x + 2 )} + \frac{x + 1}{4 ( x ^{2} + 2 x + 2 ) ^{2}}

解答补常数

= - \frac{1}{2} sin (2 arctan (x + 1)) + \frac{3 ( arctan ( x + 1 ) ( x ^{2} + 2 x + 2 ) + x + 1 )}{8 ( x ^{2} + 2 x + 2 )} + \frac{x + 1}{4 ( x ^{2} + 2 x + 2 ) ^{2}} + C

\int x cos^{3} (x^{2}) d x

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

d er i v a t i v e \frac{8}{x ^{2}}

\int \frac{24}{x ^{2} - 36} d x

\frac{\partial}{\partial x} (2^{y})