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人気のある微分積分 >

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

解

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

解

\frac{1}{4 2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

解答ステップ

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

因数

\frac{1}{x ^{4} + 4 x ^{2} + 4}

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

置換積分法を適用する

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

因数

\frac{1}{( 2 u ^{2} + 2 ) ^{2}}

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

定数を除く:

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

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

三角関数による置換を適用する

= 2 \frac{1}{4} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

三角関数の公式を使用して書き換える

= 2 \frac{1}{4} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

定数を除く:

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

= 2 \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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 \frac{1}{4} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 2 \frac{1}{4} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代用を戻す

= 2 \frac{1}{4} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

簡素化

2 \frac{1}{4} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : \frac{1}{4 2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

= \frac{1}{4 2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

定数を解答に追加する

= \frac{1}{4 2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

グラフ

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