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(t^2+t)

解

\int t^{2} + t d t

解

\frac{1}{8} (4 t t (t + 1) + 2 t (t + 1) - ln 2 t + 1 + 2 t (t + 1)) + C

解答ステップ

\int t^{2} + t d t

平方完成する

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

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

u置換積分法を適用する

= \int \frac{4 u ^{2} - 1}{2} 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 4 u^{2} - 1 d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int \frac{1}{2} tan^{2} (v) sec (v) d v

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int tan^{2} (v) sec (v) d v

三角関数の公式を使用して書き換える

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

拡張

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int - sec (v) + sec^{3} (v) d v

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}{2} \cdot \frac{1}{2} (- \int sec (v) d v + \int sec^{3} (v) d v)

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

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

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

代用を戻す

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

簡素化

\frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (t + \frac{1}{2}))) + sec (arcsec (2 (t + \frac{1}{2}))) + \frac{1}{2} sec (arcsec (2 (t + \frac{1}{2}))) tan (arcsec (2 (t + \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (2 (t + \frac{1}{2}))) + sec (arcsec (2 (t + \frac{1}{2})))) : \frac{1}{8} (4 t t (t + 1) + 2 t (t + 1) - ln 2 t + 1 + 2 t (t + 1))

= \frac{1}{8} (4 t t (t + 1) + 2 t (t + 1) - ln 2 t + 1 + 2 t (t + 1))

定数を解答に追加する

= \frac{1}{8} (4 t t (t + 1) + 2 t (t + 1) - ln 2 t + 1 + 2 t (t + 1)) + C

グラフ

人気の例

integral of x-2y

\int x - 2 y

inverselaplace-1/(s(s-1)(2s+1)(s+2))

in v erse l a pl a ce - \frac{1}{s ( s - 1 ) ( 2 s + 1 ) ( s + 2 )}

integral from 0 to 7 of 2pix(7x-x^2)

\int_{0}^{7} 2 π x (7 x - x^{2}) d x

3(dy)/(dx)+5y=1

3 \frac{d y}{d x} + 5 y = 1

derivative f(x)= 3/(\sqrt[4]{x^3)}

d er i v a t i v e f (x) = \frac{3}{4 x ^{3}}