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人気のある微分積分 >

integral of x/(sqrt(5x^2-2x+1))

解

\int \frac{x}{5 x ^{2} - 2 x + 1} d x

解

\frac{1}{5 5} (25 x^{2} - 10 x + 5 + ln (\frac{5 x - 1 + 25 x ^{2} - 10 x + 5}{2})) + C

解答ステップ

\int \frac{x}{5 x ^{2} - 2 x + 1} d x

平方完成する

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

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

u置換積分法を適用する

= \int \frac{5 u + 1}{5 25 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}{5} \cdot \int \frac{5 u + 1}{25 u ^{2} + 4} d u

拡張

\frac{5 u + 1}{25 u ^{2} + 4} : \frac{5 u}{25 u ^{2} + 4} + \frac{1}{25 u ^{2} + 4}

= \frac{1}{5} \cdot \int \frac{5 u}{25 u ^{2} + 4} + \frac{1}{25 u ^{2} + 4} d u

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}{5} (\int \frac{5 u}{25 u ^{2} + 4} d u + \int \frac{1}{25 u ^{2} + 4} d u)

\int \frac{5 u}{25 u ^{2} + 4} d u = \frac{1}{5} 25 u^{2} + 4

\int \frac{1}{25 u ^{2} + 4} d u = \frac{1}{5} ln (\frac{5 u + 4 + 25 u ^{2}}{2})

= \frac{1}{5} (\frac{1}{5} 25 u^{2} + 4 + \frac{1}{5} ln (\frac{5 u + 4 + 25 u ^{2}}{2}))

代用を戻す

u = x - \frac{1}{5}

= \frac{1}{5} \frac{1}{5} 25 (x - \frac{1}{5})^{2} + 4 + \frac{1}{5} ln \frac{5 ( x - \frac{1}{5} ) + 4 + 25 ( x - \frac{1}{5} ) ^{2}}{2}

簡素化

\frac{1}{5} \frac{1}{5} 25 (x - \frac{1}{5})^{2} + 4 + \frac{1}{5} ln \frac{5 ( x - \frac{1}{5} ) + 4 + 25 ( x - \frac{1}{5} ) ^{2}}{2} : \frac{1}{5 5} (25 x^{2} - 10 x + 5 + ln (\frac{5 x - 1 + 25 x ^{2} - 10 x + 5}{2}))

= \frac{1}{5 5} (25 x^{2} - 10 x + 5 + ln (\frac{5 x - 1 + 25 x ^{2} - 10 x + 5}{2}))

定数を解答に追加する

= \frac{1}{5 5} (25 x^{2} - 10 x + 5 + ln (\frac{5 x - 1 + 25 x ^{2} - 10 x + 5}{2})) + C

グラフ

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