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人気のある微分積分 >

integral of sqrt(16+9x^2)

解

\int 16 + 9 x^{2} d x

解

\frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4})) + C

解答ステップ

\int 16 + 9 x^{2} d x

三角関数による置換を適用する

= \int \frac{16}{3} 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{16}{3} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{16}{3} (\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{16}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{3}{4} x)

= \frac{16}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{4} x ) ) sin ( arctan ( \frac{3}{4} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x)))

簡素化

\frac{16}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{4} x ) ) sin ( arctan ( \frac{3}{4} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x))) : \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}))

= \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}))

定数を解答に追加する

= \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4})) + C

グラフ

人気の例

(e^xdx)/(-(e^x+1)^3)=(e^ydy)/((e^y+1)^2)

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}

integral of sec(5x)

\int sec (5 x) d x

inverselaplace 1/(s(s^2-2s+3))

in v erse l a pl a ce \frac{1}{s ( s ^{2} - 2 s + 3 )}

derivative ln(1/(x^2))

d er i v a t i v e ln (\frac{1}{x ^{2}})

laplacetransform 5/3

l a pl a ce t r an s f or m \frac{5}{3}