integral of x^24^{x^3}

解答

\int x^{2} 4^{x^{3}} d x

\frac{2 ^{1 + 4 x^{3}}}{3 ln ( 2 ) \cdot 4 ^{x^{3} + 1}} + C

求解步骤

\int x^{2} \cdot 4^{x^{3}} d x

使用换元积分法

= \int \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

提出常数:

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

= \frac{1}{3 ln ( 2 )} \cdot \int \frac{u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

使用分布积分法

= \frac{1}{3 ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u)

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

= \frac{1}{3 ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u))

u = 4^{x^{3}}

代回

= \frac{1}{3 ln ( 2 )} \frac{( 4 ^{x^{3}} ) ^{2}}{2 ^{\frac{l n ( 2 \cdot 4 ^{x^{3}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} \cdot 4^{x^{3}})

化简

\frac{1}{3 ln ( 2 )} \frac{( 4 ^{x^{3}} ) ^{2}}{2 ^{\frac{l n ( 2 \cdot 4 ^{x^{3}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} \cdot 4^{x^{3}}) : \frac{2 ^{1 + 4 x^{3}}}{3 ln ( 2 ) \cdot 4 ^{x^{3} + 1}}

= \frac{2 ^{1 + 4 x^{3}}}{3 ln ( 2 ) \cdot 4 ^{x^{3} + 1}}

解答补常数

= \frac{2 ^{1 + 4 x^{3}}}{3 ln ( 2 ) \cdot 4 ^{x^{3} + 1}} + C

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