ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of sqrt(4x^2-5)

解

\int 4 x^{2} - 5 d x

解

\frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

解答ステップ

\int 4 x^{2} - 5 d x

三角関数による置換を適用する

= \int \frac{5}{2} 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

= \frac{5}{2} \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

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

拡張

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

= \frac{5}{2} \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

= \frac{5}{2} (- \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) ∣

= \frac{5}{2} (- 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 (\frac{2}{5} x)

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

簡素化

\frac{5}{2} (- ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x)) + \frac{1}{2} sec (arcsec (\frac{2}{5} x)) tan (arcsec (\frac{2}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x))) : \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

定数を解答に追加する

= \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

グラフ

人気の例

integral of (2x^2)/(sqrt(7-6x^3))

integral of x^2sin^2(x^3)

integral of (4-3x)^6

integral of tcos(3t)

integral of (2x+1)(3-x)