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人気のある微分積分 >

integral of x/(sqrt(x^2+x+2))

解

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

解

x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7} + C

解答ステップ

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

平方完成する

x^{2} + x + 2 : (x + \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{7}{4}} d x

u置換積分法を適用する

= \int \frac{2 u - 1}{4 u ^{2} + 7} d u

拡張

\frac{2 u - 1}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7}

= \int \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7} 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

= \int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{1}{4 u ^{2} + 7} d u

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{2} 4 u^{2} + 7

\int \frac{1}{4 u ^{2} + 7} d u = \frac{1}{2} ln \frac{2}{7} u + \frac{1}{7} (7 + 4 u^{2})

= \frac{1}{2} 4 u^{2} + 7 - \frac{1}{2} ln \frac{2}{7} u + \frac{1}{7} (7 + 4 u^{2})

代用を戻す

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

= \frac{1}{2} 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2} ln \frac{2}{7} (x + \frac{1}{2}) + \frac{1}{7} (7 + 4 (x + \frac{1}{2})^{2})

簡素化

\frac{1}{2} 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2} ln \frac{2}{7} (x + \frac{1}{2}) + \frac{1}{7} (7 + 4 (x + \frac{1}{2})^{2}) : x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}

= x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}

定数を解答に追加する

= x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7} + C

グラフ

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