What is the integral of 1/(sqrt(1-6x^2)) ?

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

integral of 1/(sqrt(1-6x^2))

\int \frac{1}{1 - 6 x ^{2}} d x

\frac{1}{6} arcsin (6 x) + C

Solution steps

\int \frac{1}{1 - 6 x ^{2}} d x

Apply Trigonometric Substitution

= \int \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsin (6 x)

= \frac{1}{6} \cdot 1 \cdot arcsin (6 x)

Simplify

= \frac{1}{6} arcsin (6 x)

Add a constant to the solution

= \frac{1}{6} arcsin (6 x) + C

\frac{d}{d x} (\frac{x ^{3}}{x + x})

\frac{\partial}{\partial x} (1 + 2 x y)

in v erse l a pl a ce \frac{s}{( s + 2 ) ^{2} + 1}

y^{^{''}} - y^{^{'}} + 6 y = 0

\int_{2}^{\infty} \frac{1}{v ^{2} + 5 v - 6} d v

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