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

解

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

解

8 (\frac{3}{256} ln tan (\frac{1}{2} arctan (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{4}}) + C

解答ステップ

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

定数を除く:

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

= \frac{1}{4} \cdot \int \frac{x ^{4}}{x ^{2} + 1} d x

三角関数による置換を適用する

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

三角関数の公式を使用して書き換える

= \frac{1}{4} \cdot \int \frac{sin ^{4} ( u )}{cos ^{5} ( u )} d u

u置換積分法を適用する

= \frac{1}{4} \cdot \int \frac{32 v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

定数を除く:

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

= \frac{1}{4} \cdot 32 \cdot \int \frac{v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

以下の部分分数を得る:

\frac{v ^{4}}{( 1 - v ^{2} ) ^{5}} : \frac{3}{256 ( v + 1 )} + \frac{3}{256 ( v + 1 ) ^{2}} - \frac{1}{128 ( v + 1 ) ^{3}} - \frac{3}{64 ( v + 1 ) ^{4}} + \frac{1}{32 ( v + 1 ) ^{5}} - \frac{3}{256 ( v - 1 )} + \frac{3}{256 ( v - 1 ) ^{2}} + \frac{1}{128 ( v - 1 ) ^{3}} - \frac{3}{64 ( v - 1 ) ^{4}} - \frac{1}{32 ( v - 1 ) ^{5}}

= \frac{1}{4} \cdot 32 \cdot \int \frac{3}{256 ( v + 1 )} + \frac{3}{256 ( v + 1 ) ^{2}} - \frac{1}{128 ( v + 1 ) ^{3}} - \frac{3}{64 ( v + 1 ) ^{4}} + \frac{1}{32 ( v + 1 ) ^{5}} - \frac{3}{256 ( v - 1 )} + \frac{3}{256 ( v - 1 ) ^{2}} + \frac{1}{128 ( v - 1 ) ^{3}} - \frac{3}{64 ( v - 1 ) ^{4}} - \frac{1}{32 ( v - 1 ) ^{5}} 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}{4} \cdot 32 (\int \frac{3}{256 ( v + 1 )} d v + \int \frac{3}{256 ( v + 1 ) ^{2}} d v - \int \frac{1}{128 ( v + 1 ) ^{3}} d v - \int \frac{3}{64 ( v + 1 ) ^{4}} d v + \int \frac{1}{32 ( v + 1 ) ^{5}} d v - \int \frac{3}{256 ( v - 1 )} d v + \int \frac{3}{256 ( v - 1 ) ^{2}} d v + \int \frac{1}{128 ( v - 1 ) ^{3}} d v - \int \frac{3}{64 ( v - 1 ) ^{4}} d v - \int \frac{1}{32 ( v - 1 ) ^{5}} d v)

\int \frac{3}{256 ( v + 1 )} d v = \frac{3}{256} ln ∣ v + 1 ∣

\int \frac{3}{256 ( v + 1 ) ^{2}} d v = - \frac{3}{256 ( v + 1 )}

\int \frac{1}{128 ( v + 1 ) ^{3}} d v = - \frac{1}{256 ( v + 1 ) ^{2}}

\int \frac{3}{64 ( v + 1 ) ^{4}} d v = - \frac{1}{64 ( v + 1 ) ^{3}}

\int \frac{1}{32 ( v + 1 ) ^{5}} d v = - \frac{1}{128 ( v + 1 ) ^{4}}

\int \frac{3}{256 ( v - 1 )} d v = \frac{3}{256} ln ∣ v - 1 ∣

\int \frac{3}{256 ( v - 1 ) ^{2}} d v = - \frac{3}{256 ( v - 1 )}

\int \frac{1}{128 ( v - 1 ) ^{3}} d v = - \frac{1}{256 ( v - 1 ) ^{2}}

\int \frac{3}{64 ( v - 1 ) ^{4}} d v = - \frac{1}{64 ( v - 1 ) ^{3}}

\int \frac{1}{32 ( v - 1 ) ^{5}} d v = - \frac{1}{128 ( v - 1 ) ^{4}}

= \frac{1}{4} \cdot 32 (\frac{3}{256} ln ∣ v + 1 ∣ - \frac{3}{256 ( v + 1 )} - (- \frac{1}{256 ( v + 1 ) ^{2}}) - (- \frac{1}{64 ( v + 1 ) ^{3}}) - \frac{1}{128 ( v + 1 ) ^{4}} - \frac{3}{256} ln ∣ v - 1 ∣ - \frac{3}{256 ( v - 1 )} - \frac{1}{256 ( v - 1 ) ^{2}} - (- \frac{1}{64 ( v - 1 ) ^{3}}) - (- \frac{1}{128 ( v - 1 ) ^{4}}))

代用を戻す

= \frac{1}{4} \cdot 32 \frac{3}{256} ln tan (\frac{arctan ( x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arctan ( x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{4}}

簡素化

\frac{1}{4} \cdot 32 \frac{3}{256} ln tan (\frac{arctan ( x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{a r c t a n ( x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arctan ( x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{a r c t a n ( x )}{2} ) - 1 ) ^{4}} : 8 (\frac{3}{256} ln tan (\frac{1}{2} arctan (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{4}})

= 8 (\frac{3}{256} ln tan (\frac{1}{2} arctan (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{4}})

定数を解答に追加する

= 8 (\frac{3}{256} ln tan (\frac{1}{2} arctan (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( x ) ) - 1 ) ^{4}}) + C

グラフ

人気の例

面積 x^2+1,x,[-1,2]

a re a x^{2} + 1, x, [- 1, 2]

(d^2)/(dx^2)(3xe^x)

\frac{d ^{2}}{d x ^{2}} (3 x e^{x})

integral of (x-3)/((x-5)x^2)

\int \frac{x - 3}{( x - 5 ) x ^{2}} d x

勾配 (1.1)(3.4)

s l o p e (1.1) (3.4)

integral from 0 to 1 of x^3x^2

\int_{0}^{1} x^{3} x^{2} d x