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人気のある微分積分 >

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

解

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

解

\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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

\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

拡張

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

総和規則を適用する:

\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)))

代用を戻す

u = x^{2}

= \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)))

簡素化

\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)

定数を解答に追加する

= \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

グラフ

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