What is the integral of (4t)/(sqrt(1-16t^4)) ?

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

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integral of (4t)/(sqrt(1-16t^4))

\int \frac{4 t}{1 - 16 t ^{4}} d t

\frac{1}{2} arcsin (4 t^{2}) + C

Solution steps

\int \frac{4 t}{1 - 16 t ^{4}} d t

Take the constant out:

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

= 4 \cdot \int \frac{t}{1 - 16 t ^{4}} d t

Apply u-substitution

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

= 4 \cdot \int \frac{1}{2 1 - 16 u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 16 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 t^{2})

Simplify

4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 t^{2}) : \frac{1}{2} arcsin (4 t^{2})

= \frac{1}{2} arcsin (4 t^{2})

Add a constant to the solution

= \frac{1}{2} arcsin (4 t^{2}) + C

\int sin (t) sec^{5} (t) d t

x \to 0 lim ((x) x)

\int arccos (\frac{1 + x}{2}) d x

\int \frac{1}{1 + u ^{2}} d u

y = x^{2}, y = x + 6

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