What is the integral of (7x^2)/(sqrt(x^2-4)) ?

The integral of (7x^2)/(sqrt(x^2-4)) is 28(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C

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integral of (7x^2)/(sqrt(x^2-4))

\int \frac{7 x ^{2}}{x ^{2} - 4} d x

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

Solution steps

\int \frac{7 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

= 7 \cdot \int \frac{x ^{2}}{x ^{2} - 4} d x

Apply Trigonometric Substitution

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

= 7 \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

= 7 \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 7 \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) ∣

= 7 \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)

= 7 \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

7 \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))) : 28 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

= 28 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

Add a constant to the solution

= 28 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

\int x^{2} + x y^{2} + z d x

\frac{4}{x - 5}

\frac{\partial}{\partial x} (\frac{x}{y + 2 z})

\int e^{- \frac{y}{2}} d y

sin (x) + 5 π x

What is the integral of (7x^2)/(sqrt(x^2-4)) ?
The integral of (7x^2)/(sqrt(x^2-4)) is 28(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C