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

The integral of (x^3)/(sqrt(x^2-64)) is 1/3 (x^2-64)^{3/2}+64sqrt(x^2-64)+C

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

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

\frac{1}{3} (x^{2} - 64)^{\frac{3}{2}} + 64 x^{2} - 64 + C

Solution steps

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

Apply u-substitution

= \int u^{2} + 64 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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 64 d u = 64 u

= \frac{u ^{3}}{3} + 64 u

Substitute back

u = x^{2} - 64

= \frac{( x ^{2} - 64 ) ^{3}}{3} + 64 x^{2} - 64

Simplify

\frac{( x ^{2} - 64 ) ^{3}}{3} + 64 x^{2} - 64 : \frac{1}{3} (x^{2} - 64)^{\frac{3}{2}} + 64 x^{2} - 64

= \frac{1}{3} (x^{2} - 64)^{\frac{3}{2}} + 64 x^{2} - 64

Add a constant to the solution

= \frac{1}{3} (x^{2} - 64)^{\frac{3}{2}} + 64 x^{2} - 64 + C

What is the integral of (x^3)/(sqrt(x^2-64)) ?
The integral of (x^3)/(sqrt(x^2-64)) is 1/3 (x^2-64)^{3/2}+64sqrt(x^2-64)+C