integral of log_{3}(6x^2)x

解

\int lo g_{3} (6 x^{2}) x d x

\frac{1}{2} lo g_{3} (6) x^{2} + x^{2} lo g_{3} (x) - \frac{x ^{2}}{2 ln ( 3 )} + C

解答ステップ

\int lo g_{3} (6 x^{2}) x d x

部分積分を適用する

= \frac{1}{2} x^{2} lo g_{3} (6 x^{2}) - \int \frac{x}{ln ( 3 )} d x

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

= \frac{1}{2} x^{2} lo g_{3} (6 x^{2}) - \frac{x ^{2}}{2 ln ( 3 )}

簡素化

\frac{1}{2} x^{2} lo g_{3} (6 x^{2}) - \frac{x ^{2}}{2 ln ( 3 )} : \frac{1}{2} lo g_{3} (6) x^{2} + x^{2} lo g_{3} (x) - \frac{x ^{2}}{2 ln ( 3 )}

= \frac{1}{2} lo g_{3} (6) x^{2} + x^{2} lo g_{3} (x) - \frac{x ^{2}}{2 ln ( 3 )}

定数を解答に追加する

= \frac{1}{2} lo g_{3} (6) x^{2} + x^{2} lo g_{3} (x) - \frac{x ^{2}}{2 ln ( 3 )} + C