integral of 3^{sqrt(2x+1)}

解

\int 3^{2 x + 1} d x

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

解答ステップ

\int 3^{2 x + 1} d x

u置換積分法を適用する

= \int 3^{u} u d u

部分積分を適用する

= \frac{3 ^{u} u}{ln ( 3 )} - \int \frac{3 ^{u}}{ln ( 3 )} d u

\int \frac{3 ^{u}}{ln ( 3 )} d u = \frac{3 ^{u}}{ln ^{2} ( 3 )}

= \frac{3 ^{u} u}{ln ( 3 )} - \frac{3 ^{u}}{ln ^{2} ( 3 )}

代用を戻す

u = 2 x + 1

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

定数を解答に追加する

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