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人気のある微分積分 >

integral of ((x+5)/(x^2+11x+18))

解

\int (\frac{x + 5}{x ^{2} + 11 x + 18}) d x

解

2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1)) + C

解答ステップ

\int \frac{x + 5}{x ^{2} + 11 x + 18} d x

u置換積分法を適用する

= \int \frac{u}{u ^{2} + u - 12} d u

平方完成する

u^{2} + u - 12 : (u + \frac{1}{2})^{2} - \frac{49}{4}

= \int \frac{u}{( u + \frac{1}{2} ) ^{2} - \frac{49}{4}} d u

u置換積分法を適用する

= \int \frac{2 ( 2 v - 1 )}{4 v ^{2} - 49} d v

定数を除く:

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

= 2 \cdot \int \frac{2 v - 1}{4 v ^{2} - 49} d v

拡張

\frac{2 v - 1}{4 v ^{2} - 49} : \frac{2 v}{4 v ^{2} - 49} - \frac{1}{4 v ^{2} - 49}

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 v}{4 v ^{2} - 49} d v - \int \frac{1}{4 v ^{2} - 49} d v)

\int \frac{2 v}{4 v ^{2} - 49} d v = \frac{1}{4} ln 4 v^{2} - 49

\int \frac{1}{4 v ^{2} - 49} d v = - \frac{1}{28} (ln \frac{2}{7} v + 1 - ln \frac{2}{7} v - 1)

= 2 (\frac{1}{4} ln 4 v^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} v + 1 - ln \frac{2}{7} v - 1)))

代用を戻す

= 2 (\frac{1}{4} ln 4 (x + 5 + \frac{1}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + 5 + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + 5 + \frac{1}{2}) - 1)))

簡素化

2 (\frac{1}{4} ln 4 (x + 5 + \frac{1}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + 5 + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + 5 + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1))

= 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} + 44 x + 72 + \frac{1}{28} (ln \frac{2 x + 11}{7} + 1 - ln \frac{2 x + 11}{7} - 1)) + C

グラフ

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