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人気のある微分積分 >

integral of 1/(x(sqrt(4+x^2))^3)

解

\int \frac{1}{x ( 4 + x ^{2} ) ^{3}} d x

解

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

解答ステップ

\int \frac{1}{x ( 4 + x ^{2} ) ^{3}} d x

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{( u ^{2} - 4 ) u ^{2}} : - \frac{1}{16 ( u + 2 )} + \frac{1}{16 ( u - 2 )} - \frac{1}{4 u ^{2}}

= \int - \frac{1}{16 ( u + 2 )} + \frac{1}{16 ( u - 2 )} - \frac{1}{4 u ^{2}} 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

= - \int \frac{1}{16 ( u + 2 )} d u + \int \frac{1}{16 ( u - 2 )} d u - \int \frac{1}{4 u ^{2}} d u

\int \frac{1}{16 ( u + 2 )} d u = \frac{1}{16} ln ∣ u + 2 ∣

\int \frac{1}{16 ( u - 2 )} d u = \frac{1}{16} ln ∣ u - 2 ∣

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

= - \frac{1}{16} ln ∣ u + 2 ∣ + \frac{1}{16} ln ∣ u - 2 ∣ - (- \frac{1}{4 u})

代用を戻す

u = 4 + x^{2}

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

簡素化

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

定数を解答に追加する

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

グラフ

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