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人気のある微分積分 >

integral of 1/(x-sqrt(x+6))

解

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

解

ln 4 x - 4 x + 6 - \frac{1}{5} (ln \frac{2}{5} x + 6 + \frac{4}{5} - ln \frac{2}{5} x + 6 - \frac{6}{5}) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{2 u}{u ^{2} - 6 - u} 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{u}{u ^{2} - 6 - u} d u

平方完成する

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

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

u置換積分法を適用する

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

定数を除く:

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

= 2 \cdot 2 \cdot \int \frac{2 v + 1}{4 v ^{2} - 25} d v

拡張

\frac{2 v + 1}{4 v ^{2} - 25} : \frac{2 v}{4 v ^{2} - 25} + \frac{1}{4 v ^{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 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 25} d v + \int \frac{1}{4 v ^{2} - 25} d v)

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

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

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

代用を戻す

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

簡素化

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

= ln 4 x - 4 x + 6 - \frac{1}{5} (ln \frac{2}{5} x + 6 + \frac{4}{5} - ln \frac{2}{5} x + 6 - \frac{6}{5})

定数を解答に追加する

= ln 4 x - 4 x + 6 - \frac{1}{5} (ln \frac{2}{5} x + 6 + \frac{4}{5} - ln \frac{2}{5} x + 6 - \frac{6}{5}) + C

グラフ

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