integral of (1-u^2)/(u+u^3)

解答

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

ln ∣ u ∣ - ln u^{2} + 1 + C

求解步骤

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

将

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

用部份分式展开:

\frac{1}{u} - \frac{2 u}{u ^{2} + 1}

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

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}{u} d u - \int \frac{2 u}{u ^{2} + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2 u}{u ^{2} + 1} d u = ln u^{2} + 1

= ln ∣ u ∣ - ln u^{2} + 1

解答补常数

= ln ∣ u ∣ - ln u^{2} + 1 + C

\int 12 x^{2} + e^{x} d x

\frac{d}{d x} (ln (arctan (4 x^{3})))

\int 2^{x + 2} d x

\int e^{- \frac{x}{2}} cos (\frac{x}{2}) d x

l a pl a ce t r an s f or m 10 sin (3 t)