What is the integral of 1/(sqrt(1-3x^2)) ?

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

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integral of 1/(sqrt(1-3x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsin (3 x)

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial z} (5 e^{x} cos (yz))

- \frac{9}{x ^{2}}

0.8 x + 19, \frac{355.6}{x + 6}

x \to 0 lim ((1 - 7 x)^{\frac{4}{x}})

\int_{- 1}^{3} (3 x^{2} - 5) d x

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