integral of (16s-4)/(3-2s+4s^2)

解

\int \frac{16 s - 4}{3 - 2 s + 4 s ^{2}} d s

2 ln 16 s^{2} - 8 s + 12 + C

解答ステップ

\int \frac{16 s - 4}{3 - 2 s + 4 s ^{2}} d s

拡張

\frac{16 s - 4}{3 - 2 s + 4 s ^{2}} : \frac{16 s}{3 - 2 s + 4 s ^{2}} - \frac{4}{3 - 2 s + 4 s ^{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{16 s}{3 - 2 s + 4 s ^{2}} d s - \int \frac{4}{3 - 2 s + 4 s ^{2}} d s

\int \frac{16 s}{3 - 2 s + 4 s ^{2}} d s = 16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11}))

\int \frac{4}{3 - 2 s + 4 s ^{2}} d s = \frac{4}{11} arctan (\frac{4 s - 1}{11})

= 16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11})) - \frac{4}{11} arctan (\frac{4 s - 1}{11})

簡素化

16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11})) - \frac{4}{11} arctan (\frac{4 s - 1}{11}) : 2 ln 16 s^{2} - 8 s + 12

= 2 ln 16 s^{2} - 8 s + 12

定数を解答に追加する

= 2 ln 16 s^{2} - 8 s + 12 + C