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

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

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

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

arcsin (2 x^{2}) + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

= arcsin (2 x^{2})

Add a constant to the solution

= arcsin (2 x^{2}) + C

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