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

2^{x}

\frac{d}{d x} ((x + ln (x^{2}))^{s i n (x) + 2 c o s (x)})

\frac{\partial}{\partial x} (6 x^{2} y + 1)

x \to 3 lim (\frac{1}{3 x ^{2} - 27})

\frac{\partial}{\partial x} (3 sin^{2} (x) cos (x))

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