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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 64}{2 u} d u

= \int \frac{u + 64}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u + 64}{u} d u

Expand

\frac{u + 64}{u} : 1 + \frac{64}{u}

= \frac{1}{2} \cdot \int 1 + \frac{64}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u + \int \frac{64}{u} d u)

\int 1 d u = u

\int \frac{64}{u} d u = 64 ln ∣ u ∣

= \frac{1}{2} (u + 64 ln ∣ u ∣)

Substitute back

u = x^{2} - 64

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

Add a constant to the solution

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

f (x) = \frac{1}{x ^{2} + 1}, at x = 2

\frac{d}{d x} (ln (5 e^{- 4 x}))

\int \frac{x}{x ^{2} + 4 x + 3} d x

\int \frac{1}{4 - y} d y

x \to 8 + lim (\frac{x - 8}{∣ x - 8 ∣})

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