What is the integral of (x^2)/(sqrt(4x^2-9)) ?

The integral of (x^2)/(sqrt(4x^2-9)) is 1/16 (2xsqrt(4x^2-9)+9ln((|2x+sqrt(4x^2-9)|)/3))+C

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integral of (x^2)/(sqrt(4x^2-9))

\int \frac{x ^{2}}{4 x ^{2} - 9} d x

\frac{1}{16} (2 x 4 x^{2} - 9 + 9 ln (\frac{2 x + 4 x ^{2} - 9}{3})) + C

Solution steps

\int \frac{x ^{2}}{4 x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int \frac{9 sec ^{3} ( u )}{8} d u

= \int \frac{9 sec ^{3} ( u )}{8} d u

Take the constant out:

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

= \frac{9}{8} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{9}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{9}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{2}{3} x)

= \frac{9}{8} (\frac{sec ^{2} ( arcsec ( \frac{2}{3} x ) ) sin ( arcsec ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)))

Simplify

\frac{9}{8} (\frac{sec ^{2} ( arcsec ( \frac{2}{3} x ) ) sin ( arcsec ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x))) : \frac{1}{16} (2 x 4 x^{2} - 9 + 9 ln (\frac{2 x + 4 x ^{2} - 9}{3}))

= \frac{1}{16} (2 x 4 x^{2} - 9 + 9 ln (\frac{2 x + 4 x ^{2} - 9}{3}))

Add a constant to the solution

= \frac{1}{16} (2 x 4 x^{2} - 9 + 9 ln (\frac{2 x + 4 x ^{2} - 9}{3})) + C

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What is the integral of (x^2)/(sqrt(4x^2-9)) ?
The integral of (x^2)/(sqrt(4x^2-9)) is 1/16 (2xsqrt(4x^2-9)+9ln((|2x+sqrt(4x^2-9)|)/3))+C