What is the integral of ((x^3+64))/(x^2+64) ?

The integral of ((x^3+64))/(x^2+64) is 8(4((x^2)/(64)+1-ln|(x^2)/(64)+1|)+arctan(x/8))+C

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

\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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

\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

Expand

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

Apply the Sum Rule:

\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))

Substitute back

u = \frac{x}{8}

= 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}))

Add a constant to the solution

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

\frac{1}{s ( s ^{2} + 34.5 s + 1000 )}

\int_{0}^{7} 1

x \to 5 lim (x^{2} - 2 x - 15)

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

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

What is the integral of ((x^3+64))/(x^2+64) ?
The integral of ((x^3+64))/(x^2+64) is 8(4((x^2)/(64)+1-ln|(x^2)/(64)+1|)+arctan(x/8))+C