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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

因数

x^{2} + 2 x - 1 : - (- x^{2} - 2 x + 1)

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

定数を除く:

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

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

平方完成する

- x^{2} - 2 x + 1 : - (x + 1)^{2} + 2

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

u置換積分法を適用する

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

拡張

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

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

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

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

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

代用を戻す

u = x + 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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