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受欢迎的微积分 >

积分 1/((x+1)(x^2+1)^2)

解答

积分

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

解答

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

求解步骤

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

将

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

用部份分式展开:

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

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

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

化简

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} (\frac{x}{2 ( x ^{2} + 1 )} + \frac{1}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )}) : \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )}

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

解答补常数

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

作图

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