What is the integral of x/(sqrt(3-x^4)) ?

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

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integral of x/(sqrt(3-x^4))

\int \frac{x}{3 - x ^{4}} d x

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

Solution steps

\int \frac{x}{3 - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 3 - u ^{2}} d u

= \int \frac{1}{2 3 - u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{3 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

\frac{x ^{\frac{3}{2}}}{3} - x^{\frac{1}{2}}

\frac{d}{d x} (f (x) \cdot e^{\frac{1}{f ( x )}})

6 (x + 2)^{\frac{1}{2}}

\int_{0}^{6} x^{2} 36 - x^{2} d x

y = \frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}}

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