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人気のある微分積分 >

integral of (5x-3)/(x^2+4)

解

\int \frac{5 x - 3}{x ^{2} + 4} d x

解

\frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + C

解答ステップ

\int \frac{5 x - 3}{x ^{2} + 4} d x

置換積分法を適用する

= \int \frac{10 u - 3}{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}{2} \cdot \int \frac{10 u - 3}{u ^{2} + 1} d u

拡張

\frac{10 u - 3}{u ^{2} + 1} : \frac{10 u}{u ^{2} + 1} - \frac{3}{u ^{2} + 1}

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

= \frac{1}{2} (\int \frac{10 u}{u ^{2} + 1} d u - \int \frac{3}{u ^{2} + 1} d u)

\int \frac{10 u}{u ^{2} + 1} d u = 5 ln u^{2} + 1

\int \frac{3}{u ^{2} + 1} d u = 3 arctan (u)

= \frac{1}{2} (5 ln u^{2} + 1 - 3 arctan (u))

代用を戻す

u = \frac{x}{2}

= \frac{1}{2} (5 ln (\frac{x}{2})^{2} + 1 - 3 arctan (\frac{x}{2}))

(\frac{x}{2})^{2} = \frac{x ^{2}}{4}

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2}))

定数を解答に追加する

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + C

グラフ

人気の例

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