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人気のある微分積分 >

integral of (x-11)/(x^2+3x-4)

解

\int \frac{x - 11}{x ^{2} + 3 x - 4} d x

解

2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

解答ステップ

\int \frac{x - 11}{x ^{2} + 3 x - 4} d x

平方完成する

x^{2} + 3 x - 4 : (x + \frac{3}{2})^{2} - \frac{25}{4}

= \int \frac{x - 11}{( x + \frac{3}{2} ) ^{2} - \frac{25}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u - 25 )}{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

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

拡張

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

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

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

代用を戻す

u = x + \frac{3}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - (- \frac{5}{4} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1)))

簡素化

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - (- \frac{5}{4} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

グラフ

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