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

The integral of (2x)/(sqrt(1-x^2-x^4)) is arcsin(1/(sqrt(5))(2x^2+1))+C

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

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

arcsin (\frac{1}{5} (2 x^{2} + 1)) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

1 - u - u^{2} : - (u + \frac{1}{2})^{2} + \frac{5}{4}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{- ( u + \frac{1}{2} ) ^{2} + \frac{5}{4}} d u

Apply u-substitution

: \int \frac{2}{- 4 v ^{2} + 5} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{2}{- 4 v ^{2} + 5} 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 2 \cdot \int \frac{1}{- 4 v ^{2} + 5} d v

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

= 2 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{5} (x^{2} + \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{5} (x^{2} + \frac{1}{2})) : arcsin (\frac{1}{5} (2 x^{2} + 1))

= arcsin (\frac{1}{5} (2 x^{2} + 1))

Add a constant to the solution

= arcsin (\frac{1}{5} (2 x^{2} + 1)) + C

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What is the integral of (2x)/(sqrt(1-x^2-x^4)) ?
The integral of (2x)/(sqrt(1-x^2-x^4)) is arcsin(1/(sqrt(5))(2x^2+1))+C