integral of (-6x^2)/((2-2x^3)^2)

解答

\int \frac{- 6 x ^{2}}{( 2 - 2 x ^{3} ) ^{2}} d x

- 6 (- \frac{1}{36 ( x - 1 )} - \frac{1}{18 3} arctan (\frac{2 x + 1}{3}) - \frac{2}{3} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))) + C

求解步骤

\int \frac{- 6 x ^{2}}{( 2 - 2 x ^{3} ) ^{2}} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 6 \cdot \int \frac{x ^{2}}{( 2 - 2 x ^{3} ) ^{2}} d x

将

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

用部份分式展开:

\frac{1}{36 ( x - 1 ) ^{2}} - \frac{1}{36 ( x ^{2} + x + 1 )} - \frac{x}{12 ( x ^{2} + x + 1 ) ^{2}}

= - 6 \cdot \int \frac{1}{36 ( x - 1 ) ^{2}} - \frac{1}{36 ( x ^{2} + x + 1 )} - \frac{x}{12 ( x ^{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

= - 6 (\int \frac{1}{36 ( x - 1 ) ^{2}} d x - \int \frac{1}{36 ( x ^{2} + x + 1 )} d x - \int \frac{x}{12 ( x ^{2} + x + 1 ) ^{2}} d x)

\int \frac{1}{36 ( x - 1 ) ^{2}} d x = - \frac{1}{36 ( x - 1 )}

\int \frac{1}{36 ( x ^{2} + x + 1 )} d x = \frac{1}{18 3} arctan (\frac{2 x + 1}{3})

\int \frac{x}{12 ( x ^{2} + x + 1 ) ^{2}} d x = \frac{2}{3} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))

= - 6 (- \frac{1}{36 ( x - 1 )} - \frac{1}{18 3} arctan (\frac{2 x + 1}{3}) - \frac{2}{3} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))))

解答补常数

= - 6 (- \frac{1}{36 ( x - 1 )} - \frac{1}{18 3} arctan (\frac{2 x + 1}{3}) - \frac{2}{3} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))) + C

\int sin^{3} (d) x d x

t an g e n t f (x) = (x^{3} - 16 x)^{10}, (4, 0)

\int_{0}^{\frac{π}{3}} cos^{2} (x) d x

\int \frac{ln ( u )}{u} d u

l a pl a ce t r an s f or m f^{9}