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人気のある微分積分 >

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

解

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

解

4 \frac{1}{8} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{2} ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2} + C

解答ステップ

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

三角関数による置換を適用する

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

積分換算を適用する

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

代用を戻す

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

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

簡素化

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

= 4 \frac{1}{8} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{2} ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2}

定数を解答に追加する

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

グラフ

人気の例

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