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人気のある微分積分 >

integral of 10sqrt(5x^2-1)

解

\int 10 5 x^{2} - 1 d x

解

5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1 + C

解答ステップ

\int 10 5 x^{2} - 1 d x

定数を除く:

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

= 10 \cdot \int 5 x^{2} - 1 d x

三角関数による置換を適用する

= 10 \cdot \int \frac{1}{5} tan^{2} (u) sec (u) d u

定数を除く:

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

= 10 \cdot \frac{1}{5} \cdot \int tan^{2} (u) sec (u) d u

三角関数の公式を使用して書き換える

= 10 \cdot \frac{1}{5} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

拡張

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 10 \cdot \frac{1}{5} \cdot \int - sec (u) + sec^{3} (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

= 10 \cdot \frac{1}{5} (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 10 \cdot \frac{1}{5} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arcsec (5 x)

= 10 \cdot \frac{1}{5} (- ln tan (arcsec (5 x)) + sec (arcsec (5 x)) + \frac{1}{2} sec (arcsec (5 x)) tan (arcsec (5 x)) + \frac{1}{2} ln tan (arcsec (5 x)) + sec (arcsec (5 x)))

簡素化

10 \cdot \frac{1}{5} (- ln tan (arcsec (5 x)) + sec (arcsec (5 x)) + \frac{1}{2} sec (arcsec (5 x)) tan (arcsec (5 x)) + \frac{1}{2} ln tan (arcsec (5 x)) + sec (arcsec (5 x))) : 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1

= 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1

定数を解答に追加する

= 5 x 5 x^{2} - 1 - 5 ln 5 x + 5 x^{2} - 1 + C

グラフ

人気の例

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