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人気のある微分積分 >

integral of sqrt(x^2-4x+40)

解

\int x^{2} - 4 x + 40 d x

解

36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6})) + C

解答ステップ

\int x^{2} - 4 x + 40 d x

平方完成する

x^{2} - 4 x + 40 : (x - 2)^{2} + 36

= \int (x - 2)^{2} + 36 d x

u置換積分法を適用する

= \int u^{2} + 36 d u

三角関数による置換を適用する

= \int 36 sec^{3} (v) d v

定数を除く:

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

= 36 \cdot \int sec^{3} (v) d v

積分換算を適用する

= 36 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

= 36 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

代用を戻す

= 36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x - 2))) + sec (arctan (\frac{1}{6} (x - 2))))

簡素化

36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x - 2))) + sec (arctan (\frac{1}{6} (x - 2)))) : 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6}))

= 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6}))

定数を解答に追加する

= 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6})) + C

グラフ

人気の例

derivative of x^3-9x

\frac{d}{d x} (x^{3} - 9 x)

tangent f(x)= 4/x ,\at x=-2

t an g e n t f (x) = \frac{4}{x}, at x = - 2

integral from 1 to 8 of 5(ln(x))/(x^2)

\int_{1}^{8} 5 \frac{ln ( x )}{x ^{2}} d x

limit as x approaches infinity of 0.9^x

x \to \infty lim (0. 9^{x})

勾配 (5,-3),(-9,-6)

s l o p e (5, - 3), (- 9, - 6)