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人気のある微分積分 >

積分記号 x^2sqrt(a^2+x^2)

解

積分記号

x^{2} a^{2} + x^{2}

解

- \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}} + C

解答ステップ

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

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

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

三角関数による置換を適用する

= \int a^{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

= a^{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

= a^{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)

= a^{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

= a^{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) ∣)

= a^{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}{a} x)

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

簡素化

a^{4} (- (\frac{1}{2} sec (arctan (\frac{1}{a} x)) tan (arctan (\frac{1}{a} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{a} x ) ) sin ( arctan ( \frac{1}{a} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{a} x)) tan (arctan (\frac{1}{a} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x)))) : - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}}

= - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}}

定数を解答に追加する

= - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}} + C

グラフ

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