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 1/(2^x(1+4^x))

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

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

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

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

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

= \frac{1}{ln ( 2 )} (- \frac{1}{u} - arctan (u))

代用を戻す

u = 2^{x}

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

定数を解答に追加する

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

グラフ

人気の例

integral of x^2sqrt(3x+2)

\int x^{2} 3 x + 2 d x

integral of x/((x^2+2x+4)^2)

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

integral of t/((2t+3)^{5/2)}

\int \frac{t}{( 2 t + 3 ) ^{\frac{5}{2}}} d t

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

\int (cos (x))^{3} (sin (x))^{2} d x

integral of (e^{x^{-1}})/(x^2)

\int \frac{e ^{x^{- 1}}}{x ^{2}} d x