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

解

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

\frac{1}{2} ln x^{2} + 1 + C

解答ステップ

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

拡張

\frac{x ^{3} + x}{( x ^{2} + 1 ) ^{2}} : \frac{x ^{3}}{( x ^{2} + 1 ) ^{2}} + \frac{x}{( x ^{2} + 1 ) ^{2}}

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

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

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

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

簡素化

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

= \frac{1}{2} ln x^{2} + 1

定数を解答に追加する

= \frac{1}{2} ln x^{2} + 1 + C