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人気のある微分積分 >

integral of sqrt(4+25x^2)

解

\int 4 + 25 x^{2} d x

解

\frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

解答ステップ

\int 4 + 25 x^{2} d x

三角関数による置換を適用する

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

定数を除く:

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

= \frac{4}{5} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{4}{5} (\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) ∣

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

代用を戻す

u = arctan (\frac{5}{2} x)

= \frac{4}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{2} x ) ) sin ( arctan ( \frac{5}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)))

簡素化

\frac{4}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{2} x ) ) sin ( arctan ( \frac{5}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))) : \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

定数を解答に追加する

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

グラフ

人気の例

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