integral of (x+1)/(2+4x^2)

解

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

\frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x) + C

解答ステップ

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

拡張

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

\int \frac{x}{2 + 4 x ^{2}} d x = \frac{1}{8} ln 2 + 4 x^{2}

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

= \frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x)

定数を解答に追加する

= \frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x) + C