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人気のある微分積分 >

integral of x^2sqrt(x^2+1)

解

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

解

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

解答ステップ

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

三角関数による置換を適用する

= \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

= \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)

= \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

= - \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) ∣)

= - (\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 (x)

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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n or ma l - x^{3} + 2 x^{2} - 2, at x = 1

integral of-x^5e^x

\int - x^{5} e^{x} d x

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\frac{\partial}{\partial x} (4 y x^{3} - 6 y^{2} x^{2})

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t an g e n t f (x) = \frac{1}{1 + x ^{2}}, (1, \frac{1}{2})

\frac{d}{d y} (4)