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人気のある微分積分 >

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

解

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

解

\frac{3 x ^{4} arctan ( \frac{x}{2} ) + 6 x ^{3} + 24 x ^{2} arctan ( \frac{x}{2} ) + 40 x + 48 arctan ( \frac{x}{2} )}{256 ( x ^{2} + 4 ) ^{2}} + C

解答ステップ

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

指数の規則を適用する:

\frac{1}{a ^{b}} = a^{- b}

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

= \int (x^{2} + 4)^{- 3} d x

置換積分法を適用する

= \int \frac{1}{32 ( u ^{2} + 1 ) ^{3}} d u

定数を除く:

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

= \frac{1}{32} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{3}} d u

積分換算を適用する

= \frac{1}{32} (\frac{3}{4} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

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

= \frac{1}{32} (\frac{3}{4} \cdot \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

代用を戻す

u = \frac{x}{2}

= \frac{1}{32} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{\frac{x}{2}}{1 + ( \frac{x}{2} ) ^{2}}) + \frac{\frac{x}{2}}{2 \cdot 2 ( ( \frac{x}{2} ) ^{2} + 1 ) ^{2}}

簡素化

\frac{1}{32} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{\frac{x}{2}}{1 + ( \frac{x}{2} ) ^{2}}) + \frac{\frac{x}{2}}{2 \cdot 2 ( ( \frac{x}{2} ) ^{2} + 1 ) ^{2}} : \frac{3 x ^{4} arctan ( \frac{x}{2} ) + 6 x ^{3} + 24 x ^{2} arctan ( \frac{x}{2} ) + 40 x + 48 arctan ( \frac{x}{2} )}{256 ( x ^{2} + 4 ) ^{2}}

= \frac{3 x ^{4} arctan ( \frac{x}{2} ) + 6 x ^{3} + 24 x ^{2} arctan ( \frac{x}{2} ) + 40 x + 48 arctan ( \frac{x}{2} )}{256 ( x ^{2} + 4 ) ^{2}}

定数を解答に追加する

= \frac{3 x ^{4} arctan ( \frac{x}{2} ) + 6 x ^{3} + 24 x ^{2} arctan ( \frac{x}{2} ) + 40 x + 48 arctan ( \frac{x}{2} )}{256 ( x ^{2} + 4 ) ^{2}} + C

グラフ

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