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

解答

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

- \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )} + C

求解步骤

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

将

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

用部份分式展开:

- \frac{3}{16 ( 2 x + 1 )} - \frac{9}{32 ( 2 x + 1 ) ^{2}} + \frac{5}{16 ( 2 x - 1 )} - \frac{7}{32 ( 2 x - 1 ) ^{2}}

= \int - \frac{3}{16 ( 2 x + 1 )} - \frac{9}{32 ( 2 x + 1 ) ^{2}} + \frac{5}{16 ( 2 x - 1 )} - \frac{7}{32 ( 2 x - 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{3}{16 ( 2 x + 1 )} d x - \int \frac{9}{32 ( 2 x + 1 ) ^{2}} d x + \int \frac{5}{16 ( 2 x - 1 )} d x - \int \frac{7}{32 ( 2 x - 1 ) ^{2}} d x

\int \frac{3}{16 ( 2 x + 1 )} d x = \frac{3}{32} ln ∣ 2 x + 1 ∣

\int \frac{9}{32 ( 2 x + 1 ) ^{2}} d x = - \frac{9}{64 ( 2 x + 1 )}

\int \frac{5}{16 ( 2 x - 1 )} d x = \frac{5}{32} ln ∣ 2 x - 1 ∣

\int \frac{7}{32 ( 2 x - 1 ) ^{2}} d x = - \frac{7}{64 ( 2 x - 1 )}

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ - (- \frac{9}{64 ( 2 x + 1 )}) + \frac{5}{32} ln ∣ 2 x - 1 ∣ - (- \frac{7}{64 ( 2 x - 1 )})

化简

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )}

解答补常数

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )} + C

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

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

in t e g r a l \frac{1}{x ^{2} - 1}

\int - cos (x) 2 d x

\int \frac{s}{s + 1} d s