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

= \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

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

y = x^{3} - 2 x + 1, (3, 22)

(- 3 x + 14)^{\frac{4}{3}}, 2

\frac{\partial}{\partial x} (x^{5} + x^{2} y + x + 1)

\int \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

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