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

解答

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

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

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

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

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

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

化简

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

解答补常数

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

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

\int cot^{- 4} (d) x d x

\int (1 + tan^{2} (5 x)) d x

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

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