integral of ((6x^3+6x))/(x^2+1)

解

\int \frac{( 6 x ^{3} + 6 x )}{x ^{2} + 1} d x

3 x^{2} + 3 + C

解答ステップ

\int \frac{6 x ^{3} + 6 x}{x ^{2} + 1} d x

拡張

\frac{6 x ^{3} + 6 x}{x ^{2} + 1} : \frac{6 x ^{3}}{x ^{2} + 1} + \frac{6 x}{x ^{2} + 1}

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 \frac{6 x ^{3}}{x ^{2} + 1} d x + \int \frac{6 x}{x ^{2} + 1} d x

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

\int \frac{6 x}{x ^{2} + 1} d x = 3 ln x^{2} + 1

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

簡素化

3 (x^{2} + 1 - ln x^{2} + 1) + 3 ln x^{2} + 1 : 3 x^{2} + 3

= 3 x^{2} + 3

定数を解答に追加する

= 3 x^{2} + 3 + C