What is the integral of (t^2)/(sqrt(1-t^6)) ?

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

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integral of (t^2)/(sqrt(1-t^6))

\int \frac{t ^{2}}{1 - t ^{6}} d t

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

Solution steps

\int \frac{t ^{2}}{1 - t ^{6}} d t

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = arcsin (u)

= \frac{1}{3} arcsin (u)

Substitute back

u = t^{3}

= \frac{1}{3} arcsin (t^{3})

Add a constant to the solution

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

\int e^{t^{2}} t^{3} d t

\frac{d ^{2}}{d x ^{2}} (4 x^{3} - 5 x^{4})

sin (x y) = cos (x + y)

\frac{( 1 0 ^{x} )}{ln ( 10 )}

\frac{\partial}{\partial x} (20 - 4 x^{2} - 3 y^{2})

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