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

解答

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

\frac{1}{2} ln x^{2} + 1 - 2 arctan (x) - \frac{1}{2} ln 4 x^{2} - 4 x + 4 + \frac{5}{3} arctan (\frac{2 x - 1}{3}) + C

求解步骤

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

将

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

用部份分式展开:

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

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

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

\int \frac{- x + 3}{x ^{2} - x + 1} d x = - \frac{1}{2} ln 4 x^{2} - 4 x + 4 + \frac{5}{3} arctan (\frac{2 x - 1}{3})

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

解答补常数

= \frac{1}{2} ln x^{2} + 1 - 2 arctan (x) - \frac{1}{2} ln 4 x^{2} - 4 x + 4 + \frac{5}{3} arctan (\frac{2 x - 1}{3}) + C

\int e^{4 x^{2}} 8 x d x

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

\int \frac{2}{200 + t} d t

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\int \frac{6 x - 3}{4 x ^{2} - 4 x + 1} d x