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人気のある微分積分 >

integral of (sqrt(1-x^4))/(x^5)

解

\int \frac{1 - x ^{4}}{x ^{5}} d x

解

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

解答ステップ

\int \frac{1 - x ^{4}}{x ^{5}} d x

u置換積分法を適用する

= \int \frac{1 - u ^{2}}{2 u ^{3}} d u

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{1 - u ^{2}}{u ^{3}} d u

部分積分を適用する

= \frac{1}{2} (- \frac{1 - u ^{2}}{2 u ^{2}} - \int \frac{1}{2 u - u ^{2} + 1} d u)

\int \frac{1}{2 u - u ^{2} + 1} d u = - \frac{1}{4} (ln - u^{2} + 1 + 1 - ln - u^{2} + 1 - 1)

= \frac{1}{2} (- \frac{1 - u ^{2}}{2 u ^{2}} - (- \frac{1}{4} (ln - u^{2} + 1 + 1 - ln - u^{2} + 1 - 1)))

代用を戻す

u = x^{2}

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

簡素化

\frac{1}{2} - \frac{1 - ( x ^{2} ) ^{2}}{2 ( x ^{2} ) ^{2}} - (- \frac{1}{4} (ln - (x^{2})^{2} + 1 + 1 - ln - (x^{2})^{2} + 1 - 1)) : \frac{1}{2} (- \frac{1 - x ^{4}}{2 x ^{4}} + \frac{1}{4} (ln - x^{4} + 1 + 1 - ln - x^{4} + 1 - 1))

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

定数を解答に追加する

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

グラフ

人気の例

integral of (6x^4+x-3)/(2x^3+2x^2-12x)

\int \frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} d x

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\int \frac{7 x ^{2}}{( x ^{2} + 3 ) ^{2}} d x

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