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人気のある微分積分 >

integral of (9x^2)/(sqrt(x^2-4))

解

\int \frac{9 x ^{2}}{x ^{2} - 4} d x

解

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

解答ステップ

\int \frac{9 x ^{2}}{x ^{2} - 4} d x

定数を除く:

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

= 9 \cdot \int \frac{x ^{2}}{x ^{2} - 4} d x

三角関数による置換を適用する

= 9 \cdot \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

= 9 \cdot 4 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 9 \cdot 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) ∣

= 9 \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

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

= 9 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)))

簡素化

9 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : 36 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

= 36 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

定数を解答に追加する

= 36 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

グラフ

人気の例

x y^{^{'}} = y^{2} - y

integral from 0 to 7 of sqrt(x^2-49)

\int_{0}^{7} x^{2} - 49 d x

(d^2)/(dx^2)(sqrt(x^2+64))

\frac{d ^{2}}{d x ^{2}} (x^{2} + 64)

integral of 7ln(x^2-x+8)

\int 7 ln (x^{2} - x + 8) d x

derivative of ax^3+bx^2+c

\frac{d}{d x} (a x^{3} + b x^{2} + c)