integral of 2xsec^2(3x)x

解

\int 2 x sec^{2} (3 x) x d x

2 (\frac{1}{3} x^{2} tan (3 x) - \frac{i}{27} Li_{2} (- e^{2 \cdot 3 i x}) - \frac{i x ^{2}}{3} + \frac{2}{9} x ln (1 + cos (6 x) + i sin (6 x))) + C

解答ステップ

\int 2 x sec^{2} (3 x) x d x

定数を除く:

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

= 2 \cdot \int x sec^{2} (3 x) x d x

簡素化

xx : x^{2}

= 2 \cdot \int sec^{2} (3 x) x^{2} d x

部分積分を適用する

= 2 (\frac{1}{3} x^{2} tan (3 x) - \int \frac{2}{3} x tan (3 x) d x)

\int \frac{2}{3} x tan (3 x) d x = \frac{i}{27} Li_{2} (- e^{2 \cdot 3 i x}) + \frac{i x ^{2}}{3} - \frac{2}{9} x ln (1 + cos (6 x) + i sin (6 x))

= 2 (\frac{1}{3} x^{2} tan (3 x) - (\frac{i}{27} Li_{2} (- e^{2 \cdot 3 i x}) + \frac{i x ^{2}}{3} - \frac{2}{9} x ln (1 + cos (6 x) + i sin (6 x))))

簡素化

= 2 (\frac{1}{3} x^{2} tan (3 x) - \frac{i}{27} Li_{2} (- e^{2 \cdot 3 i x}) - \frac{i x ^{2}}{3} + \frac{2}{9} x ln (1 + cos (6 x) + i sin (6 x)))

定数を解答に追加する

= 2 (\frac{1}{3} x^{2} tan (3 x) - \frac{i}{27} Li_{2} (- e^{2 \cdot 3 i x}) - \frac{i x ^{2}}{3} + \frac{2}{9} x ln (1 + cos (6 x) + i sin (6 x))) + C