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人気のある微分積分 >

integral of e^{2x}5^{2x}

解

\int e^{2 x} 5^{2 x} d x

解

\frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}} + C

解答ステップ

\int e^{2 x} \cdot 5^{2 x} d x

5^{2 x} = 2 5^{x}

= \int 2 5^{x} e^{2 x} d x

u置換積分法を適用する

= \int \frac{u ^{\frac{1}{l n ( 5 )}}}{2 ln ( 5 )} d u

定数を除く:

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

= \frac{1}{2 ln ( 5 )} \cdot \int u^{\frac{1}{l n ( 5 )}} d u

べき乗則を適用する

= \frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) u^{\frac{1 + l n ( 5 )}{l n ( 5 )}}

代用を戻す

u = 2 5^{x}

= \frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) (2 5^{x})^{\frac{1 + l n ( 5 )}{l n ( 5 )}}

簡素化

\frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) (2 5^{x})^{\frac{1 + l n ( 5 )}{l n ( 5 )}} : \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}}

= \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}}

定数を解答に追加する

= \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}} + C

グラフ

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