integral of 1/(16+25x^2)+2/(3(x-2))

解

\int \frac{1}{16 + 25 x ^{2}} + \frac{2}{3 ( x - 2 )} d x

\frac{1}{20} arctan (\frac{5}{4} x) + \frac{2}{3} ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{1}{16 + 25 x ^{2}} + \frac{2}{3 ( x - 2 )} d x

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{1}{16 + 25 x ^{2}} d x + \int \frac{2}{3 ( x - 2 )} d x

\int \frac{1}{16 + 25 x ^{2}} d x = \frac{1}{20} arctan (\frac{5}{4} x)

\int \frac{2}{3 ( x - 2 )} d x = \frac{2}{3} ln ∣ x - 2 ∣

= \frac{1}{20} arctan (\frac{5}{4} x) + \frac{2}{3} ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{1}{20} arctan (\frac{5}{4} x) + \frac{2}{3} ln ∣ x - 2 ∣ + C