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

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

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integral of (2x)/(sqrt(1-4x^4))

\int \frac{2 x}{1 - 4 x ^{4}} d x

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

Solution steps

\int \frac{2 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

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

Apply u-substitution

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

= 2 \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

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

Apply Trigonometric Substitution

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

= 2 \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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{z ^{2}}{z ^{3} + 1} d z

- 9 (5 x^{2} + 4)^{- 6}

x + y = x^{3} + y^{2}

\int e^{t a n (x)} sec^{2} (x) d x

y = 4 x^{2}

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