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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

置換積分法を適用する

= \int \frac{2 ( 2 u ^{3} + 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

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

長除法

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

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

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

\int 2 u d u = u^{2}

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

= 2 (u^{2} - ln u^{2} + 1 + arctan (u))

代用を戻す

u = \frac{x}{2}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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