integral of (2x+1)/(4x^2+4x+7)

解

\int \frac{2 x + 1}{4 x ^{2} + 4 x + 7} d x

\frac{1}{4} ln 4 x^{2} + 4 x + 7 + C

解答ステップ

\int \frac{2 x + 1}{4 x ^{2} + 4 x + 7} d x

拡張

\frac{2 x + 1}{4 x ^{2} + 4 x + 7} : \frac{2 x}{4 x ^{2} + 4 x + 7} + \frac{1}{4 x ^{2} + 4 x + 7}

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{2 x}{4 x ^{2} + 4 x + 7} d x + \int \frac{1}{4 x ^{2} + 4 x + 7} d x

\int \frac{2 x}{4 x ^{2} + 4 x + 7} d x = \frac{1}{4} ln 4 x^{2} + 4 x + 7 - \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1))

\int \frac{1}{4 x ^{2} + 4 x + 7} d x = \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1))

= \frac{1}{4} ln 4 x^{2} + 4 x + 7 - \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1)) + \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1))

類似した元を足す:

- \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1)) + \frac{1}{2 6} arctan (\frac{1}{6} (2 x + 1)) = 0

= \frac{1}{4} ln 4 x^{2} + 4 x + 7

定数を解答に追加する

= \frac{1}{4} ln 4 x^{2} + 4 x + 7 + C