What is the integral of 1/(x^3sqrt(4x^2-1)) ?

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

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integral of 1/(x^3sqrt(4x^2-1))

\int \frac{1}{x ^{3} 4 x ^{2} - 1} d x

2 arcsec (2 x) + sin (2 arcsec (2 x)) + C

Solution steps

\int \frac{1}{x ^{3} 4 x ^{2} - 1} d x

Apply Trigonometric Substitution

: \int 4 cos^{2} (u) d u

= \int 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (2 x)

= 4 \cdot \frac{1}{2} (arcsec (2 x) + \frac{1}{2} sin (2 arcsec (2 x)))

Simplify

4 \cdot \frac{1}{2} (arcsec (2 x) + \frac{1}{2} sin (2 arcsec (2 x))) : 2 arcsec (2 x) + sin (2 arcsec (2 x))

= 2 arcsec (2 x) + sin (2 arcsec (2 x))

Add a constant to the solution

= 2 arcsec (2 x) + sin (2 arcsec (2 x)) + C

\int \frac{1}{3 y ^{2}} d y

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

\int (x + 1)^{2} sin (x) d x

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

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