integral from 1 to x^5 of log_{3}(t+2)

解

\int_{1}^{x^{5}} lo g_{3} (t + 2) d t

\frac{x ^{5} ln ( x ^{5} + 2 ) - x ^{5} + 2 ln x ^{5} + 2 + 1 - 3 ln ( 3 )}{ln ( 3 )}

解答ステップ

\int_{1}^{x^{5}} lo g_{3} (t + 2) d t

部分積分を適用する

= [t lo g_{3} (t + 2) - \int \frac{t}{ln ( 3 ) ( t + 2 )} d t]_{1}^{x^{5}}

\int \frac{t}{ln ( 3 ) ( t + 2 )} d t = \frac{1}{ln ( 3 )} (t + 2 - 2 ln ∣ t + 2 ∣)

= [t lo g_{3} (t + 2) - \frac{1}{ln ( 3 )} (t + 2 - 2 ln ∣ t + 2 ∣)]_{1}^{x^{5}}

結合

t lo g_{3} (t + 2) - \frac{1}{ln ( 3 )} (t + 2 - 2 ln ∣ t + 2 ∣) : \frac{- t - 2 + 2 ln ∣ t + 2 ∣ + t ln ( t + 2 )}{ln ( 3 )}

= [\frac{- t - 2 + 2 ln ∣ t + 2 ∣ + t ln ( t + 2 )}{ln ( 3 )}]_{1}^{x^{5}}

限度を計算する:

\frac{- x ^{5} - 2 + 2 ln x ^{5} + 2 + x ^{5} ln ( x ^{5} + 2 )}{ln ( 3 )} - \frac{3 ln ( 3 ) - 3}{ln ( 3 )}

= \frac{- x ^{5} - 2 + 2 ln x ^{5} + 2 + x ^{5} ln ( x ^{5} + 2 )}{ln ( 3 )} - \frac{3 ln ( 3 ) - 3}{ln ( 3 )}

簡素化

= \frac{x ^{5} ln ( x ^{5} + 2 ) - x ^{5} + 2 ln x ^{5} + 2 + 1 - 3 ln ( 3 )}{ln ( 3 )}