integral of (5x^3-3x^2+2x-1)/(x^4+x^2)

解

\int \frac{5 x ^{3} - 3 x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} d x

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

解答ステップ

\int \frac{5 x ^{3} - 3 x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} d x

拡張

\frac{5 x ^{3} - 3 x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} : \frac{5 x ^{3}}{x ^{4} + x ^{2}} - \frac{3 x ^{2}}{x ^{4} + x ^{2}} + \frac{2 x}{x ^{4} + x ^{2}} - \frac{1}{x ^{4} + x ^{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{5 x ^{3}}{x ^{4} + x ^{2}} d x - \int \frac{3 x ^{2}}{x ^{4} + x ^{2}} d x + \int \frac{2 x}{x ^{4} + x ^{2}} d x - \int \frac{1}{x ^{4} + x ^{2}} d x

\int \frac{5 x ^{3}}{x ^{4} + x ^{2}} d x = \frac{5}{2} ln x^{2} + 1

\int \frac{3 x ^{2}}{x ^{4} + x ^{2}} d x = 3 arctan (x)

\int \frac{2 x}{x ^{4} + x ^{2}} d x = - ln x^{2} + 1 + 2 ln (x)

\int \frac{1}{x ^{4} + x ^{2}} d x = - \frac{1}{x} - arctan (x)

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

簡素化

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

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

定数を解答に追加する

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