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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

t \to 2 - lim (- \frac{3 t}{t - 2})

x \to 1 + lim (\frac{3}{x ^{2} - 1})

11 x^{2} e^{x}

4 x^{2}

n = 0 \sum \infty \frac{e ^{n}}{( n ! )}

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