What is the integral of t/(sqrt(1-t^4)) ?

The integral of t/(sqrt(1-t^4)) is 1/2 arcsin(t^2)+C

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integral of t/(sqrt(1-t^4))

\int \frac{t}{1 - t ^{4}} d t

\frac{1}{2} arcsin (t^{2}) + C

Solution steps

\int \frac{t}{1 - t ^{4}} d t

Apply u-substitution

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

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

Substitute back

u = t^{2}

= \frac{1}{2} arcsin (t^{2})

Add a constant to the solution

= \frac{1}{2} arcsin (t^{2}) + C

h = (x x^{2} + 2), (x + h, (x + h)^{2} + 2)

\int \frac{2 x}{x ^{2} - 27} d x

\int \frac{3}{2} x d x

y^{^{'}} = y sin (4 x)

\frac{d}{d y} (2 x sin (y))

What is the integral of t/(sqrt(1-t^4)) ?
The integral of t/(sqrt(1-t^4)) is 1/2 arcsin(t^2)+C