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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

三角関数による置換を適用する

= \int \frac{9 sec ^{3} ( u )}{2 2} d u

定数を除く:

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

= \frac{9}{2 2} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{9}{2 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) ∣

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

代用を戻す

u = arcsec (\frac{2}{3} x)

= \frac{9}{2 2} \frac{sec ^{2} ( arcsec ( \frac{2}{3} x ) ) sin ( arcsec ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x))

簡素化

\frac{9}{2 2} \frac{sec ^{2} ( arcsec ( \frac{2}{3} x ) ) sin ( arcsec ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)) : \frac{1}{4} x 2 x^{2} - 9 + \frac{9}{4 2} ln (\frac{2 x ^{2} - 9 + 2 x}{3})

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

定数を解答に追加する

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

グラフ

人気の例

limit as x approaches-1 of 7x+2

x \to - 1 lim (7 x + 2)

integral from-6 to 8 of x/2+8

\int_{- 6}^{8} \frac{x}{2} + 8 d x

integral of x^5sqrt(5+x^6)

\int x^{5} 5 + x^{6} d x

integral of (x^3-6x+8)/(x^2)

\int \frac{x ^{3} - 6 x + 8}{x ^{2}} d x

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

\int \frac{2}{( x ^{2} - 1 )} d x