What is the integral of (2x)/(sqrt(-5-6x^2-x^4)) ?

The integral of (2x)/(sqrt(-5-6x^2-x^4)) is arcsin(1/2 (x^2+3))+C

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integral of (2x)/(sqrt(-5-6x^2-x^4))

\int \frac{2 x}{- 5 - 6 x ^{2} - x ^{4}} d x

arcsin (\frac{1}{2} (x^{2} + 3)) + C

Solution steps

\int \frac{2 x}{- 5 - 6 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}{- 5 - 6 x ^{2} - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 - 5 - 6 u - u ^{2}} d u

= 2 \cdot \int \frac{1}{2 - 5 - 6 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}{- 5 - 6 u - u ^{2}} d u

Complete the square

- 5 - 6 u - u^{2} : - (u + 3)^{2} + 4

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{- ( u + 3 ) ^{2} + 4} d u

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d w

= 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 1 \cdot w

Substitute back

= 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} (x^{2} + 3))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} (x^{2} + 3)) : arcsin (\frac{1}{2} (x^{2} + 3))

= arcsin (\frac{1}{2} (x^{2} + 3))

Add a constant to the solution

= arcsin (\frac{1}{2} (x^{2} + 3)) + C

\frac{7 x}{8 - tan ( x )}

\int \frac{4 x ^{2} + x + 4}{( x ^{2} + 1 ) ^{2}} d x

\int_{a}^{b} (\frac{1}{x ^{3}} + 3 x) d x

\int - \frac{1}{1 + sin ( x )} d x

x y^{^{'}} - 2 y = x^{4} - 4

What is the integral of (2x)/(sqrt(-5-6x^2-x^4)) ?
The integral of (2x)/(sqrt(-5-6x^2-x^4)) is arcsin(1/2 (x^2+3))+C