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人気のある微分積分 >

integral of (2x+1)/(x^2-x-6)

解

\int \frac{2 x + 1}{x ^{2} - x - 6} d x

解

ln 4 x^{2} - 4 x - 24 - \frac{2}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + C

解答ステップ

\int \frac{2 x + 1}{x ^{2} - x - 6} d x

平方完成する

x^{2} - x - 6 : (x - \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{2 x + 1}{( x - \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

u置換積分法を適用する

= \int \frac{4 ( 2 u + 2 )}{4 u ^{2} - 25} d u

定数を除く:

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

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

拡張

\frac{2 u + 2}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{2}{4 u ^{2} - 25}

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

= 4 (\int \frac{2 u}{4 u ^{2} - 25} d u + \int \frac{2}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{2}{4 u ^{2} - 25} d u = - \frac{1}{10} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

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

代用を戻す

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

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

簡素化

4 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 25 - \frac{1}{10} (ln \frac{2}{5} (x - \frac{1}{2}) + 1 - ln \frac{2}{5} (x - \frac{1}{2}) - 1)) : ln 4 x^{2} - 4 x - 24 - \frac{2}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5})

= ln 4 x^{2} - 4 x - 24 - \frac{2}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5})

定数を解答に追加する

= ln 4 x^{2} - 4 x - 24 - \frac{2}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + C

グラフ

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