What is the integral of (x^2)/(sqrt(16-x^6)) ?

The integral of (x^2)/(sqrt(16-x^6)) is 1/3 arcsin(1/4 x^3)+C

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integral of (x^2)/(sqrt(16-x^6))

\int \frac{x ^{2}}{16 - x ^{6}} d x

\frac{1}{3} arcsin (\frac{1}{4} x^{3}) + C

Solution steps

\int \frac{x ^{2}}{16 - x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 16 - u ^{2}} d u

= \int \frac{1}{3 16 - u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{1}{4} x^{3})

Simplify

= \frac{1}{3} arcsin (\frac{1}{4} x^{3})

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{1}{4} x^{3}) + C

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

y^{^{'}} = \frac{x ^{3} + y ^{3}}{x y ^{2}}

\frac{d y}{d x} y = 11 x

f (x) = - x^{4} + 4 x^{3} + 8 x^{2} + 5, at x = 2

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

What is the integral of (x^2)/(sqrt(16-x^6)) ?
The integral of (x^2)/(sqrt(16-x^6)) is 1/3 arcsin(1/4 x^3)+C