integral of (4x^2+x)/((16x^3+6x^2)^2)

解答

\int \frac{4 x ^{2} + x}{( 16 x ^{3} + 6 x ^{2} ) ^{2}} d x

\frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )} + C

求解步骤

\int \frac{4 x ^{2} + x}{( 16 x ^{3} + 6 x ^{2} ) ^{2}} d x

将

\frac{4 x ^{2} + x}{( 16 x ^{3} + 6 x ^{2} ) ^{2}}

用部份分式展开:

- \frac{1}{27 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{64}{27 ( 8 x + 3 ) ^{2}}

= \int - \frac{1}{27 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{64}{27 ( 8 x + 3 ) ^{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}{27 x ^{2}} d x + \int \frac{1}{36 x ^{3}} d x + \int \frac{64}{27 ( 8 x + 3 ) ^{2}} d x

\int \frac{1}{27 x ^{2}} d x = - \frac{1}{27 x}

\int \frac{1}{36 x ^{3}} d x = - \frac{1}{72 x ^{2}}

\int \frac{64}{27 ( 8 x + 3 ) ^{2}} d x = - \frac{8}{27 ( 8 x + 3 )}

= - (- \frac{1}{27 x}) - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )}

化简

= \frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )}

解答补常数

= \frac{1}{27 x} - \frac{1}{72 x ^{2}} - \frac{8}{27 ( 8 x + 3 )} + C

\int \frac{( e ^{3 x} )}{1 + e ^{6 x}} d x

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

\int 6 x cos (2 x) d x

\int x^{2} y e^{- y x} d x

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