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人気のある微分積分 >

integral of 6x^24^{x^3}

解

\int 6 x^{2} 4^{x^{3}} d x

解

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

解答ステップ

\int 6 x^{2} \cdot 4^{x^{3}} d x

定数を除く:

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

= 6 \cdot \int x^{2} \cdot 4^{x^{3}} d x

u置換積分法を適用する

= 6 \cdot \int \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

定数を除く:

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

= 6 \cdot \frac{1}{3 ln ( 2 )} \cdot \int \frac{u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

部分積分を適用する

= 6 \cdot \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

= 6 \cdot \frac{1}{3 ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u))

代用を戻す

u = 4^{x^{3}}

= 6 \cdot \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}})

簡素化

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

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

定数を解答に追加する

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

グラフ

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