What is the integral of x^3sqrt(1+64x^2) ?

The integral of x^3sqrt(1+64x^2) is 1/30720 (96x^2-1)(64x^2+1)^{3/2}+C

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integral of x^3sqrt(1+64x^2)

\int x^{3} 1 + 64 x^{2} d x

\frac{1}{30720} (96 x^{2} - 1) (64 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 + 64 x^{2} d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{8192} d u

= \int \frac{u ( u - 1 )}{8192} d u

Take the constant out:

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

= \frac{1}{8192} \cdot \int u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{8192} \cdot \int u^{\frac{3}{2}} - 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}{8192} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{1}{8192} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + 64 x^{2}

= \frac{1}{8192} (\frac{2}{5} (1 + 64 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 64 x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{8192} (\frac{2}{5} (1 + 64 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 64 x^{2})^{\frac{3}{2}}) : \frac{1}{30720} (96 x^{2} - 1) (64 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{30720} (96 x^{2} - 1) (64 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{30720} (96 x^{2} - 1) (64 x^{2} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (16 x^{2})

\frac{d}{d x} (\frac{3 x ^{2}}{2} - \frac{7}{5 x ^{2}})

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

y = 3 x^{2} - 4 x + 5

y = ln (x) x

What is the integral of x^3sqrt(1+64x^2) ?
The integral of x^3sqrt(1+64x^2) is 1/30720 (96x^2-1)(64x^2+1)^{3/2}+C