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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{4 5} (3 ln \frac{2 x ^{2} - 1}{5} + 1 - 3 ln \frac{2 x ^{2} - 1}{5} - 1 - 5 ln - 4 x^{4} + 4 x^{2} + 4) + C

求解步骤

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

使用换元积分法

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

提出常数:

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

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

乘开

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

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

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

\int \frac{1}{u + 1 - u ^{2}} d u = \frac{1}{5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1)

\int \frac{u}{u + 1 - u ^{2}} d u = 2 (- \frac{1}{4} ln - 4 u^{2} + 4 u + 4 + \frac{1}{4 5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1))

= \frac{1}{2} (\frac{1}{5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1) + 2 (- \frac{1}{4} ln - 4 u^{2} + 4 u + 4 + \frac{1}{4 5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1)))

u = x^{2}

代回

= \frac{1}{2} (\frac{1}{5} (ln \frac{2 x ^{2} - 1}{5} + 1 - ln \frac{2 x ^{2} - 1}{5} - 1) + 2 (- \frac{1}{4} ln - 4 (x^{2})^{2} + 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} - 1}{5} + 1 - ln \frac{2 x ^{2} - 1}{5} - 1)))

化简

\frac{1}{2} (\frac{1}{5} (ln \frac{2 x ^{2} - 1}{5} + 1 - ln \frac{2 x ^{2} - 1}{5} - 1) + 2 (- \frac{1}{4} ln - 4 (x^{2})^{2} + 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} - 1}{5} + 1 - ln \frac{2 x ^{2} - 1}{5} - 1))) : \frac{1}{4 5} (3 ln \frac{2 x ^{2} - 1}{5} + 1 - 3 ln \frac{2 x ^{2} - 1}{5} - 1 - 5 ln - 4 x^{4} + 4 x^{2} + 4)

= \frac{1}{4 5} (3 ln \frac{2 x ^{2} - 1}{5} + 1 - 3 ln \frac{2 x ^{2} - 1}{5} - 1 - 5 ln - 4 x^{4} + 4 x^{2} + 4)

解答补常数

= \frac{1}{4 5} (3 ln \frac{2 x ^{2} - 1}{5} + 1 - 3 ln \frac{2 x ^{2} - 1}{5} - 1 - 5 ln - 4 x^{4} + 4 x^{2} + 4) + C

作图

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