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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

三角関数による置換を適用する

= \int 2 sec^{3} (u) d u

定数を除く:

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

= 2 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 2 (\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) ∣

= 2 (\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)

= 2 \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

2 \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)) : \frac{1}{2} x x^{2} - 2 + ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x

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

定数を解答に追加する

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

グラフ

人気の例

derivative y=(3x)/(sqrt(1-x))

d er i v a t i v e y = \frac{3 x}{1 - x}

(\partial)/(\partial x)(cosh(y)+x-2)

\frac{\partial}{\partial x} (cosh (y) + x - 2)

integral from pi/4 to pi of (sin(x))

\int_{\frac{π}{4}}^{π} (sin (x)) d x

integral from 4 to infinity of e^{5x}

\int_{4}^{\infty} e^{5 x} d x

integral of (2+x^2)/(1+x^2)

\int \frac{2 + x ^{2}}{1 + x ^{2}} d x