What is the integral of (21x^2)/(sqrt(x^2-4)) ?

The integral of (21x^2)/(sqrt(x^2-4)) is 84(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C

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integral of (21x^2)/(sqrt(x^2-4))

\int \frac{21 x ^{2}}{x ^{2} - 4} d x

84 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

Solution steps

\int \frac{21 x ^{2}}{x ^{2} - 4} d x

Take the constant out:

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

= 21 \cdot \int \frac{x ^{2}}{x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int 4 sec^{3} (u) d u

= 21 \cdot \int 4 sec^{3} (u) d u

Take the constant out:

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

= 21 \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 21 \cdot 4 (\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) ∣

= 21 \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{2} x)

= 21 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)))

Simplify

21 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : 84 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

= 84 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

Add a constant to the solution

= 84 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

\frac{d}{d x} (\frac{3 x ^{5} + 8}{x ^{3}})

n = 1 \sum \infty \frac{n}{n ^{2} + 4}

\int 1 - y d y

\int 2 x ln (2 x) d x

y^{^{''}} - 6 y^{^{'}} + 10 y = 0

What is the integral of (21x^2)/(sqrt(x^2-4)) ?
The integral of (21x^2)/(sqrt(x^2-4)) is 84(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C