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人気のある微分積分 >

integral of (10x)/(x^2+x-6)

解

\int \frac{10 x}{x ^{2} + x - 6} d x

解

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

解答ステップ

\int \frac{10 x}{x ^{2} + x - 6} d x

定数を除く:

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

= 10 \cdot \int \frac{x}{x ^{2} + x - 6} d x

平方完成する

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

= 10 \cdot \int \frac{x}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

u置換積分法を適用する

= 10 \cdot \int \frac{2 ( 2 u - 1 )}{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

= 10 \cdot 2 \cdot \int \frac{2 u - 1}{4 u ^{2} - 25} d u

拡張

\frac{2 u - 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} - \frac{1}{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

= 10 \cdot 2 (\int \frac{2 u}{4 u ^{2} - 25} d u - \int \frac{1}{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{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 10 \cdot 2 (\frac{1}{4} ln 4 u^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

代用を戻す

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

= 10 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

簡素化

10 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 5 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x

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

定数を解答に追加する

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

グラフ

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