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

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

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

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

\frac{4 x ^{2} - 1}{x} + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int cos (v) d v

= 2 \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 2 sin (v)

Substitute back

= 2 sin (arcsec (2 x))

Simplify

2 sin (arcsec (2 x)) : \frac{4 x ^{2} - 1}{x}

= \frac{4 x ^{2} - 1}{x}

Add a constant to the solution

= \frac{4 x ^{2} - 1}{x} + C

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

f (x) = 2 x^{2} + x - 1, at a = - 2

\int e^{x} x cos (x) d x

f (x) = 1 - 2 x

x \to 6 lim (x^{3} + 4 x)

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