What is the integral of x/(sqrt(9-x^4)) ?

The integral of x/(sqrt(9-x^4)) is 1/2 arcsin(1/3 x^2)+C

integral of x/(sqrt(9-x^4))

\int \frac{x}{9 - x ^{4}} d x

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

Solution steps

\int \frac{x}{9 - x ^{4}} d x

Apply u-substitution

= \int \frac{1}{2 9 - 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}{9 - u ^{2}} d u

Apply Trigonometric Substitution

= \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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

n \to \infty lim (\frac{1}{x ^{n - 1}})

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

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

d er i v a t i v e \frac{1}{3 x ^{2} - 1}

What is the integral of x/(sqrt(9-x^4)) ?
The integral of x/(sqrt(9-x^4)) is 1/2 arcsin(1/3 x^2)+C