What is the integral of (x^2)/((x^2-1)^{3/2)} ?

The integral of (x^2)/((x^2-1)^{3/2)} is -x/((x^2-1)^{1/2)}+ln|(x^2-1)^{1/2}+x|+C

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integral of (x^2)/((x^2-1)^{3/2)}

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec^{3} (u) cot^{2} (u) d u

= \int sec^{3} (u) cot^{2} (u) d u

Rewrite using trig identities

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

Apply Integration By Parts

: - csc (u) - \int - sec (u) d u

= - csc (u) - \int - sec (u) d u

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

= - csc (u) - (- ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (x)

= - csc (arcsec (x)) - (- ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣)

Simplify

- csc (arcsec (x)) - (- ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣) : - \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}} + ln (x^{2} - 1)^{\frac{1}{2}} + x

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

Add a constant to the solution

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

f (x) = x^{2} - 5 x, at x = - 2

\int 9 x sin (x) cos (x) d x

f (x) = ∣ x ∣ 5 - x^{2}, (2, 2)

x \to 2 lim (3 x + 4 x)

\int_{2}^{4} \frac{2}{x x ^{2} - 4} d x

What is the integral of (x^2)/((x^2-1)^{3/2)} ?
The integral of (x^2)/((x^2-1)^{3/2)} is -x/((x^2-1)^{1/2)}+ln|(x^2-1)^{1/2}+x|+C