integral of (4^u+ln(5^u))

解

\int (4^{u} + ln (5^{u})) d u

\frac{2 ^{2 u - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) u^{2} + C

解答ステップ

\int 4^{u} + ln (5^{u}) d u

以下のように仮定して対数の規則

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

を適用する:

x \geq 0

= \int 4^{u} + ln (5) u d u

総和規則を適用する:

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

= \int 4^{u} d u + \int ln (5) u d u

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

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

= \frac{2 ^{2 u - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) u^{2}

定数を解答に追加する

= \frac{2 ^{2 u - 1}}{ln ( 2 )} + \frac{1}{2} ln (5) u^{2} + C