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人気のある微分積分 >

integral of x^2sqrt(x^2+2)

解

\int x^{2} x^{2} + 2 d x

解

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

解答ステップ

\int x^{2} x^{2} + 2 d x

三角関数による置換を適用する

= \int 4 sec^{3} (u) tan^{2} (u) d u

定数を除く:

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

= 4 \cdot \int sec^{3} (u) tan^{2} (u) d u

三角関数の公式を使用して書き換える

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

拡張

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 4 \cdot \int - sec^{3} (u) + sec^{5} (u) 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

= 4 (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= 4 (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

代用を戻す

u = arctan (\frac{1}{2} x)

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

簡素化

4 - (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{2} x ) ) sin ( arctan ( \frac{1}{2} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))) : - \frac{1}{4} x 2 + x^{2} - \frac{1}{2} ln \frac{1}{2} x + \frac{1}{2} (2 + x^{2}) + \frac{1}{4} x (x^{2} + 2)^{\frac{3}{2}}

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

定数を解答に追加する

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

グラフ

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