integral of 1/(16x^4-1)

解答

\int \frac{1}{16 x ^{4} - 1} d x

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

求解步骤

\int \frac{1}{16 x ^{4} - 1} d x

将

\frac{1}{16 x ^{4} - 1}

用部份分式展开:

- \frac{1}{2 ( 4 x ^{2} + 1 )} - \frac{1}{4 ( 2 x + 1 )} + \frac{1}{4 ( 2 x - 1 )}

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{4 ( 2 x + 1 )} d x = \frac{1}{8} ln ∣ 2 x + 1 ∣

\int \frac{1}{4 ( 2 x - 1 )} d x = \frac{1}{8} ln ∣ 2 x - 1 ∣

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

解答补常数

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

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

\frac{d y}{d x} = sin (17 x)

\int (x + 1) \cdot e^{4 x + 1} d x

d er i v a t i v e 3 4 x - 5

\int 3 x^{2} - x d x