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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用三角换元法

= \int 16 tan^{4} (u) sec (u) d u

提出常数:

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

= 16 \cdot \int tan^{4} (u) sec (u) d u

使用三角恒等式改写

= 16 \cdot \int \frac{sin ^{4} ( u )}{cos ^{5} ( u )} d u

使用换元积分法

= 16 \cdot \int \frac{32 v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

提出常数:

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

= 16 \cdot 32 \cdot \int \frac{v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

将

\frac{v ^{4}}{( 1 - v ^{2} ) ^{5}}

用部份分式展开:

\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}}

= 16 \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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \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}}

= 16 \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}}))

代回

= 16 \cdot 32 \frac{3}{256} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{4}}

化简

16 \cdot 32 \frac{3}{256} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{a r c t a n ( \frac{1}{2} x )}{2} ) - 1 ) ^{4}} : 512 (\frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{4}})

= 512 (\frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{4}})

解答补常数

= 512 (\frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arctan (\frac{1}{2} x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arctan ( \frac{1}{2} x ) ) - 1 ) ^{4}}) + C

作图

流行的例子

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\int_{- 2}^{3} x - 4 ∣ x ∣ d x

t a y l or \frac{1}{x ^{2}}

derivative of 6^{3x-4}

\frac{d}{d x} (6^{3 x - 4})

integral from 0 to x of sin(pi/2 t^2)

\int_{0}^{x} sin (\frac{π}{2} t^{2}) d t

derivative of 2^x-x^2

\frac{d}{d x} (2^{x} - x^{2})