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人気のある微分積分 >

integral of 1/(x^3sqrt(x^2-a^2))

解

\int \frac{1}{x ^{3} x ^{2} - a ^{2}} d x

解

\frac{1}{2 a ^{3}} (arcsec (\frac{x}{a}) + \frac{1}{2} sin (2 arcsec (\frac{x}{a}))) + C

解答ステップ

\int \frac{1}{x ^{3} x ^{2} - a ^{2}} d x

三角関数による置換を適用する

= \int \frac{1}{a ^{3} sec ^{2} ( u )} d u

定数を除く:

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

= \frac{1}{a ^{3}} \cdot \int \frac{1}{sec ^{2} ( u )} d u

三角関数の公式を使用して書き換える

= \frac{1}{a ^{3}} \cdot \int \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= \frac{1}{a ^{3}} \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a ^{3}} \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{1}{a ^{3}} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

代用を戻す

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

= \frac{1}{a ^{3}} \cdot \frac{1}{2} (arcsec (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{a} x)))

簡素化

\frac{1}{a ^{3}} \cdot \frac{1}{2} (arcsec (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsec (\frac{1}{a} x))) : \frac{1}{2 a ^{3}} (arcsec (\frac{x}{a}) + \frac{1}{2} sin (2 arcsec (\frac{x}{a})))

= \frac{1}{2 a ^{3}} (arcsec (\frac{x}{a}) + \frac{1}{2} sin (2 arcsec (\frac{x}{a})))

定数を解答に追加する

= \frac{1}{2 a ^{3}} (arcsec (\frac{x}{a}) + \frac{1}{2} sin (2 arcsec (\frac{x}{a}))) + C

グラフ

人気の例

integral of 150x

\int 150 x d x

integral of sin(41x)sin(cos(41x))

\int sin (41 x) sin (cos (41 x)) d x

integral of 1/(sqrt(5x+9))

\int \frac{1}{5 x + 9} d x

integral of (x^8-2x^4)^3(x^7-x^3)

\int (x^{8} - 2 x^{4})^{3} (x^{7} - x^{3}) d x

integral of 1/((64x^2+1)^2)

\int \frac{1}{( 64 x ^{2} + 1 ) ^{2}} d x