What is the integral of 1/(sqrt(1-4x^2)) ?

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

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integral of 1/(sqrt(1-4x^2))

\int \frac{1}{1 - 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{1 - 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

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

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsin (2 x)

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d r} (\frac{( r ^{2} - 1 ) ^{3}}{( 2 r + 1 ) ^{5}})

\int_{4}^{6} (x^{2} + 2 x - 8) d x

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

\int \frac{25}{x} d x

\frac{d}{d x} (\frac{x ^{2} - 5 x}{x})

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