integral of 1/((x^2+1)^3)

解答

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

\frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} + C

求解步骤

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

使用降次积分法

= \frac{3}{4} \cdot \int \frac{1}{( x ^{2} + 1 ) ^{2}} d x + \frac{x}{2 \cdot 2 ( x ^{2} + 1 ) ^{2}}

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

= \frac{3}{4} \cdot \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{2 \cdot 2 ( x ^{2} + 1 ) ^{2}}

化简

\frac{3}{4} \cdot \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{2 \cdot 2 ( x ^{2} + 1 ) ^{2}} : \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}}

= \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}}

解答补常数

= \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} + C

\frac{\partial}{\partial x} (ln (x^{2} + x y))

y^{^{''}} + y^{^{'}} - 5 x = 0

\int_{- \infty}^{x} \frac{1}{x ^{2}} d x

t an g e n t y = 2 + 1 x + 10 x e^{x}, at x = 0

y^{^{''}} + y = cos (2 x)