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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d v

= 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 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y = 2 x - x^{2}, (2, 0)

\int_{- 1}^{1} x^{- 3} d x

\frac{\partial}{\partial x} (\frac{e ^{x} - e ^{y}}{e ^{x y}})

\frac{\partial}{\partial u} (2 cos (u))

y = - 2 x^{2} - 4 x + 1, (0, 1)

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