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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 x^{2})

Simplify

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

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

Add a constant to the solution

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

y = \frac{x}{a} + \frac{a}{x}, at x = \frac{a}{2}

x \to 5 lim ([x])

f (x) = x^{2} + 8, (- 3, 17)

x \to - 2 lim (4 x)

y^{^{'}} - 2 (x + 1)^{3} = \frac{4}{( x + 1 )} y

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