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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 4 - u ^{2}} d u

= \int \frac{1}{2 4 - 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}{4 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

Add a constant to the solution

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

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