integral of 1/(9x^4+x^2)

解答

\int \frac{1}{9 x ^{4} + x ^{2}} d x

- \frac{1}{x} - 3 arctan (3 x) + C

求解步骤

\int \frac{1}{9 x ^{4} + x ^{2}} d x

将

\frac{1}{9 x ^{4} + x ^{2}}

用部份分式展开:

\frac{1}{x ^{2}} - \frac{9}{9 x ^{2} + 1}

= \int \frac{1}{x ^{2}} - \frac{9}{9 x ^{2} + 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}{x ^{2}} d x - \int \frac{9}{9 x ^{2} + 1} d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{9}{9 x ^{2} + 1} d x = 3 arctan (3 x)

= - \frac{1}{x} - 3 arctan (3 x)

解答补常数

= - \frac{1}{x} - 3 arctan (3 x) + C

\int \frac{5}{x ^{\frac{1}{2}}} d x

\int \frac{3 x}{1 + 9 x ^{2}} d x

\int \frac{4}{3 x ^{2}} d x

\int 5 csc (x) cot (x) d x

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