integral of 1/((4-x^2)^{1.5)}

解

\int \frac{1}{( 4 - x ^{2} ) ^{1.5}} d x

- \frac{0.03125 x ^{5} - 0.25 x ^{3} + 0.5 x}{4 ( 0.5 x + 1 ) ( 0.5 x - 1 ) ( - 0.25 x ^{2} + 1 ) ^{\frac{3}{2}}} + C

解答ステップ

\int \frac{1}{( 4 - x ^{2} ) ^{1.5}} d x

(4 - x^{2})^{1.5} = (4 - x^{2}) 4 - x^{2}

= \int \frac{1}{( 4 - x ^{2} ) 4 - x ^{2}} d x

4 - x^{2} = 4 (1 - \frac{x ^{2}}{4})

= \int \frac{1}{4 ( 1 - \frac{x ^{2}}{4} ) 4 - x ^{2}} d x

定数を除く:

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

= \frac{1}{4} \cdot \int \frac{1}{( 1 - \frac{x ^{2}}{4} ) 4 - x ^{2}} d x

4 - x^{2} = 4 (1 - \frac{x ^{2}}{4})

= \frac{1}{4} \cdot \int \frac{1}{( 1 - \frac{x ^{2}}{4} ) 4 ( 1 - \frac{x ^{2}}{4} )} d x

部分積分を適用する

= \frac{1}{4} (\frac{0.5 x}{( 1 - 0.25 x ^{2} ) 1 - 0.25 x ^{2}} - \int \frac{0.375 x ^{2}}{( - 0.25 x ^{2} + 1 ) ^{\frac{5}{2}}} d x)

\int \frac{0.375 x ^{2}}{( - 0.25 x ^{2} + 1 ) ^{\frac{5}{2}}} d x = \frac{0.125 x ^{3}}{( 1 - 0.25 x ^{2} ) ^{\frac{3}{2}}}

= \frac{1}{4} (\frac{0.5 x}{( 1 - 0.25 x ^{2} ) 1 - 0.25 x ^{2}} - \frac{0.125 x ^{3}}{( 1 - 0.25 x ^{2} ) ^{\frac{3}{2}}})

簡素化

\frac{1}{4} (\frac{0.5 x}{( 1 - 0.25 x ^{2} ) 1 - 0.25 x ^{2}} - \frac{0.125 x ^{3}}{( 1 - 0.25 x ^{2} ) ^{\frac{3}{2}}}) : - \frac{0.03125 x ^{5} - 0.25 x ^{3} + 0.5 x}{4 ( 0.5 x + 1 ) ( 0.5 x - 1 ) ( - 0.25 x ^{2} + 1 ) ^{\frac{3}{2}}}

= - \frac{0.03125 x ^{5} - 0.25 x ^{3} + 0.5 x}{4 ( 0.5 x + 1 ) ( 0.5 x - 1 ) ( - 0.25 x ^{2} + 1 ) ^{\frac{3}{2}}}

定数を解答に追加する

= - \frac{0.03125 x ^{5} - 0.25 x ^{3} + 0.5 x}{4 ( 0.5 x + 1 ) ( 0.5 x - 1 ) ( - 0.25 x ^{2} + 1 ) ^{\frac{3}{2}}} + C