What is the integral of (x^2)/(sqrt(x^2-25)) ?

The integral of (x^2)/(sqrt(x^2-25)) is 25(1/50 xsqrt(x^2-25)+1/2 ln((|sqrt(x^2-25)+x|)/5))+C

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integral of (x^2)/(sqrt(x^2-25))

\int \frac{x ^{2}}{x ^{2} - 25} d x

25 (\frac{1}{50} x x^{2} - 25 + \frac{1}{2} ln (\frac{x ^{2} - 25 + x}{5})) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 25} d x

Apply Trigonometric Substitution

: \int 25 sec^{3} (u) d u

= \int 25 sec^{3} (u) d u

Take the constant out:

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

= 25 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 25 (\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) ∣

= 25 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{5} x)

= 25 (\frac{sec ^{2} ( arcsec ( \frac{1}{5} x ) ) sin ( arcsec ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)))

Simplify

25 (\frac{sec ^{2} ( arcsec ( \frac{1}{5} x ) ) sin ( arcsec ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))) : 25 (\frac{1}{50} x x^{2} - 25 + \frac{1}{2} ln (\frac{x ^{2} - 25 + x}{5}))

= 25 (\frac{1}{50} x x^{2} - 25 + \frac{1}{2} ln (\frac{x ^{2} - 25 + x}{5}))

Add a constant to the solution

= 25 (\frac{1}{50} x x^{2} - 25 + \frac{1}{2} ln (\frac{x ^{2} - 25 + x}{5})) + C

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

\frac{d}{d x} ((x^{3} - x + 1)^{10})

y = \frac{1}{x ^{4}}, (1, 1)

\int \frac{x ^{6}}{( 3 - x ^{7} ) ^{2}} d x

\int \frac{- 6}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

What is the integral of (x^2)/(sqrt(x^2-25)) ?
The integral of (x^2)/(sqrt(x^2-25)) is 25(1/50 xsqrt(x^2-25)+1/2 ln((|sqrt(x^2-25)+x|)/5))+C