integral of ((x^2))/(x^2+64)

解答

\int \frac{( x ^{2} )}{x ^{2} + 64} d x

- 8 arctan (\frac{x}{8}) + x + C

求解步骤

\int \frac{x ^{2}}{x ^{2} + 64} d x

\frac{x ^{2}}{x ^{2} + 64} = - \frac{64}{x ^{2} + 64} + 1

= \int - \frac{64}{x ^{2} + 64} + 1 d x

使用换元积分法

= \int 8 (- \frac{1}{u ^{2} + 1} + 1) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

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

= 8 (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= 8 (- arctan (u) + u)

u = \frac{x}{8}

代回

= 8 (- arctan (\frac{x}{8}) + \frac{x}{8})

化简

8 (- arctan (\frac{x}{8}) + \frac{x}{8}) : - 8 arctan (\frac{x}{8}) + x

= - 8 arctan (\frac{x}{8}) + x

解答补常数

= - 8 arctan (\frac{x}{8}) + x + C

\frac{\partial}{\partial x} (- 3 sin (3 x))

t an g e n t 4 x - \frac{x ^{3}}{4}

\int - y sin (y) d y

t a y l or e^{\frac{1}{x - 9}}, 7

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