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人気のある微分積分 >

integral of x/(4x^2-4x+5)

解

\int \frac{x}{4 x ^{2} - 4 x + 5} d x

解

\frac{1}{8} (ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})) + C

解答ステップ

\int \frac{x}{4 x ^{2} - 4 x + 5} d x

平方完成する

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

= \int \frac{x}{4 ( x - \frac{1}{2} ) ^{2} + 4} d x

u置換積分法を適用する

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

拡張

\frac{2 u + 1}{4 u ^{2} + 4} : \frac{2 u}{4 u ^{2} + 4} + \frac{1}{4 u ^{2} + 4}

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{2 u}{4 u ^{2} + 4} d u + \int \frac{1}{4 u ^{2} + 4} d u)

\int \frac{2 u}{4 u ^{2} + 4} d u = \frac{1}{4} ln 4 u^{2} + 4

\int \frac{1}{4 u ^{2} + 4} d u = \frac{1}{4} arctan (u)

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

代用を戻す

u = x - \frac{1}{2}

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

簡素化

\frac{1}{2} (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 4 + \frac{1}{4} arctan (x - \frac{1}{2})) : \frac{1}{8} (ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))

= \frac{1}{8} (ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))

定数を解答に追加する

= \frac{1}{8} (ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})) + C

グラフ

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