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(1+(4x)^2)

解

\int 1 + (4 x)^{2} d x

解

\frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2}) + C

解答ステップ

\int 1 + (4 x)^{2} d x

簡素化

1 + (4 x)^{2} : 1 + 16 x^{2}

= \int 1 + 16 x^{2} d x

三角関数による置換を適用する

= \int \frac{1}{4} sec^{3} (u) d u

定数を除く:

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

= \frac{1}{4} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{1}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= \frac{1}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (4 x)

= \frac{1}{4} (\frac{sec ^{2} ( arctan ( 4 x ) ) sin ( arctan ( 4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣)

簡素化

\frac{1}{4} (\frac{sec ^{2} ( arctan ( 4 x ) ) sin ( arctan ( 4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣) : \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2})

= \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2})

定数を解答に追加する

= \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2}) + C

グラフ

人気の例

integral of 1/(sin(x)cos^2(x))

\int \frac{1}{sin ( x ) cos ^{2} ( x )} d x

tangent f(x)=10sin(x)+10sqrt(3)cos(x)

t an g e n t f (x) = 10 sin (x) + 103 cos (x)

inverselaplace 2/(s^2+3s+2)

in v erse l a pl a ce \frac{2}{s ^{2} + 3 s + 2}

sum from n=5 to infinity of (n^2)/(n!)

n = 5 \sum \infty \frac{n ^{2}}{n !}

integral from-1 to 1 of sqrt(1+x^{-2/3)}

\int_{- 1}^{1} 1 + x^{- \frac{2}{3}} d x