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

解答

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

\frac{2}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + C

求解步骤

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

将

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

用部份分式展开:

\frac{2}{9 ( x + 1 )} + \frac{1}{3 ( x + 1 ) ^{2}} + \frac{- 2 x - 1}{9 ( x ^{2} + 2 )}

= \int \frac{2}{9 ( x + 1 )} + \frac{1}{3 ( x + 1 ) ^{2}} + \frac{- 2 x - 1}{9 ( x ^{2} + 2 )} 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{2}{9 ( x + 1 )} d x + \int \frac{1}{3 ( x + 1 ) ^{2}} d x + \int \frac{- 2 x - 1}{9 ( x ^{2} + 2 )} d x

\int \frac{2}{9 ( x + 1 )} d x = \frac{2}{9} ln ∣ x + 1 ∣

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

\int \frac{- 2 x - 1}{9 ( x ^{2} + 2 )} d x = \frac{1}{9} (- ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2}))

= \frac{2}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2}))

解答补常数

= \frac{2}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + C

\int 1 - cos (\frac{1}{x}) d x

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

\int \frac{2}{x ^{3} - 8} d x

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

\int tan^{- 3} (x) sec^{2} (x) d x