integral of (3x-1)/(x^2+10x+29)

解

\int \frac{3 x - 1}{x ^{2} + 10 x + 29} d x

3 (\frac{1}{2} ln x^{2} + 10 x + 29 - \frac{5}{2} arctan (\frac{1}{2} (x + 5))) - \frac{1}{2} arctan (\frac{x + 5}{2}) + C

解答ステップ

\int \frac{3 x - 1}{x ^{2} + 10 x + 29} d x

拡張

\frac{3 x - 1}{x ^{2} + 10 x + 29} : \frac{3 x}{x ^{2} + 10 x + 29} - \frac{1}{x ^{2} + 10 x + 29}

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{3 x}{x ^{2} + 10 x + 29} d x - \int \frac{1}{x ^{2} + 10 x + 29} d x

\int \frac{3 x}{x ^{2} + 10 x + 29} d x = 3 (\frac{1}{2} ln x^{2} + 10 x + 29 - \frac{5}{2} arctan (\frac{1}{2} (x + 5)))

\int \frac{1}{x ^{2} + 10 x + 29} d x = \frac{1}{2} arctan (\frac{x + 5}{2})

= 3 (\frac{1}{2} ln x^{2} + 10 x + 29 - \frac{5}{2} arctan (\frac{1}{2} (x + 5))) - \frac{1}{2} arctan (\frac{x + 5}{2})

定数を解答に追加する

= 3 (\frac{1}{2} ln x^{2} + 10 x + 29 - \frac{5}{2} arctan (\frac{1}{2} (x + 5))) - \frac{1}{2} arctan (\frac{x + 5}{2}) + C