integral of ((x^3+64))/(x^2+64)

解答

\int \frac{( x ^{3} + 64 )}{x ^{2} + 64} d x

8 (4 (\frac{x ^{2}}{64} + 1 - ln \frac{x ^{2}}{64} + 1) + arctan (\frac{x}{8})) + C

求解步骤

\int \frac{x ^{3} + 64}{x ^{2} + 64} d x

使用换元积分法

= \int \frac{8 ( 8 u ^{3} + 1 )}{u ^{2} + 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{8 u ^{3} + 1}{u ^{2} + 1} d u

乘开

\frac{8 u ^{3} + 1}{u ^{2} + 1} : \frac{8 u ^{3}}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

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{8 u ^{3}}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u)

\int \frac{8 u ^{3}}{u ^{2} + 1} d u = 4 (u^{2} + 1 - ln u^{2} + 1)

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

= 8 (4 (u^{2} + 1 - ln u^{2} + 1) + arctan (u))

u = \frac{x}{8}

代回

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

4 ((\frac{x}{8})^{2} + 1 - ln (\frac{x}{8})^{2} + 1) = 4 (\frac{x ^{2}}{64} + 1 - ln \frac{x ^{2}}{64} + 1)

= 8 (4 (\frac{x ^{2}}{64} + 1 - ln \frac{x ^{2}}{64} + 1) + arctan (\frac{x}{8}))

解答补常数

= 8 (4 (\frac{x ^{2}}{64} + 1 - ln \frac{x ^{2}}{64} + 1) + arctan (\frac{x}{8})) + C

\int_{0}^{7} 1

d er i v a t i v e (x (x^{2} + 4))^{\frac{1}{3}}

\frac{\partial}{\partial x} (6 + 3 x^{2} - 2 y^{2})

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

l a pl a ce t r an s f or m 2 t^{4}