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人気のある微分積分 >

integral of 1/(2x^2-16)

解

\int \frac{1}{2 x ^{2} - 16} d x

解

- \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

解答ステップ

\int \frac{1}{2 x ^{2} - 16} d x

置換積分法を適用する

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

定数を除く:

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

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

因数

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

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

定数を除く:

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

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

共通積分を使用する:

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

= \frac{1}{4 2} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

代用を戻す

u = \frac{2}{4} x

= \frac{1}{4 2} - \frac{ln \frac{2}{4} x + 1}{2} - \frac{ln \frac{2}{4} x - 1}{2}

簡素化

\frac{1}{4 2} - \frac{ln \frac{2}{4} x + 1}{2} - \frac{ln \frac{2}{4} x - 1}{2} : - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

= - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

定数を解答に追加する

= - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

グラフ

人気の例

integral of (sqrt(4-x^2))/x

\int \frac{4 - x ^{2}}{x} d x

limit as x approaches-2 of x/((x+2)^2)

x \to - 2 lim (\frac{x}{( x + 2 ) ^{2}})

limit as x approaches-4 of (x)^2

x \to - 4 lim ((x)^{2})

integral of 1/(sqrt(x^2+3))

\int \frac{1}{x ^{2} + 3} d x

limit as θ approaches 0 of θcos(3/θ)

θ \to 0 lim (θ cos (\frac{3}{θ}))