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

The integral of x^3sqrt(64+x^2) is 1/5 (64+x^2)^{5/2}-64/3 (64+x^2)^{3/2}+C

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integral of x^3sqrt(64+x^2)

\int x^{3} 64 + x^{2} d x

\frac{1}{5} (64 + x^{2})^{\frac{5}{2}} - \frac{64}{3} (64 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 64 + x^{2} d x

Apply u-substitution

: \int (u^{2} - 64) u^{2} d u

= \int (u^{2} - 64) u^{2} d u

Expand

(u^{2} - 64) u^{2} : u^{4} - 64 u^{2}

= \int u^{4} - 64 u^{2} 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^{4} d u - \int 64 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 64 u^{2} d u = \frac{64 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{64 u ^{3}}{3}

Substitute back

u = 64 + x^{2}

= \frac{( 64 + x ^{2} ) ^{5}}{5} - \frac{64 ( 64 + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 64 + x ^{2} ) ^{5}}{5} - \frac{64 ( 64 + x ^{2} ) ^{3}}{3} : \frac{1}{5} (64 + x^{2})^{\frac{5}{2}} - \frac{64}{3} (64 + x^{2})^{\frac{3}{2}}

= \frac{1}{5} (64 + x^{2})^{\frac{5}{2}} - \frac{64}{3} (64 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (64 + x^{2})^{\frac{5}{2}} - \frac{64}{3} (64 + x^{2})^{\frac{3}{2}} + C

\frac{d ^{6}}{d x ^{6}} (sin (x^{2}))

x \to 0 lim (\frac{x}{sin ( 7 x )})

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

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

f (t) = t e^{5 - 2 t}

What is the integral of x^3sqrt(64+x^2) ?
The integral of x^3sqrt(64+x^2) is 1/5 (64+x^2)^{5/2}-64/3 (64+x^2)^{3/2}+C