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人気のある微分積分 >

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

解

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

解

- \frac{1}{5} (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{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{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}{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{1}{4 u ^{2} - 25} d u

置換積分法を適用する

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

定数を除く:

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

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

因数

v^{2} - 1 : - (- v^{2} + 1)

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 \cdot \frac{1}{10} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代用を戻す

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

簡素化

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

= - \frac{1}{5} (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

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

グラフ

人気の例

t an g e n t 9 x^{2}

derivative of ln(x^2-5x+2)

\frac{d}{d x} (ln (x^{2} - 5 x + 2))

integral of (ln(2))/(x^2)

\int \frac{ln ( 2 )}{x ^{2}} d x

y^'=((3t^2-e^t))/((2y-5)),y(0)=1

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{( 2 y - 5 )}, y (0) = 1

sum from n=1 to infinity of ln(n+2)

n = 1 \sum \infty ln (n + 2)