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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

= \int - \frac{1}{- u ^{2} + 16} d u

定数を除く:

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

= - \int \frac{1}{- u ^{2} + 16} d u

置換積分法を適用する

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

定数を除く:

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

= - \frac{1}{4} \cdot \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}

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

代用を戻す

= - \frac{1}{4} \frac{ln \frac{16 - x ^{2}}{4} + 1}{2} - \frac{ln \frac{16 - x ^{2}}{4} - 1}{2}

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

d er i v a t i v e f^{9}

integral of (7x^3)/(sqrt(4+x^2))

\int \frac{7 x ^{3}}{4 + x ^{2}} d x

tangent y=-2x^3,(1,-2)

t an g e n t y = - 2 x^{3}, (1, - 2)

integral of (3x^2)

\int (3 x^{2}) d x

(\partial)/(\partial x)(sin^2(3xy^2-y))

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))