Help with calculus (integrals).?

When you integrate, include the differential (in this case, dx)

sqrt(x^2 - 1) * dx / x^3

Now we can use a substitution

x = sec(t)

dx = sec(t) * tan(t) * dt

sqrt(sec(t)^2 - 1) * sec(t) * tan(t) * dt / sec(t)^3

sqrt(tan(t)^2) * tan(t) * dt / sec(t)^2

tan(t)^2 * dt / sec(t)^2

(sec(t)^2 - 1) * dt / sec(t)^2

sec(t)^2 * dt / sec(t)^2 - dt / sec(t)^2

dt - cos(t)^2 * dt

dt - (1/2) * (1 + cos(2t)) * dt =>

dt - (1/2) * dt - (1/2) * cos(2t) * dt =>

(1/2) * dt - (1/2) * cos(2t) * dt

Integrate

(1/2) * t - (1/4) * sin(2t) + C

(1/2) * t - (1/2) * sin(t) * cos(t) + C

(1/2) * t - (1/2) * sin(t) * cos(t)^2 / cos(t) + C

(1/2) * t - (1/2) * tan(t) * cos(t)^2 + C

(1/2) * t - (1/2) * tan(t) / sec(t)^2 + C

(1/2) * t - (1/2) * sqrt(sec(t)^2 - 1) / sec(t)^2 + C

(1/2) * (t - sqrt(sec(t)^2 - 1) / sec(t)^2) + C

(1/2) * (arcsec(x) - sqrt(x^2 - 1) / x^2) + C

Now we apply the limits from -7 to -2

(1/2) * (arcsec(-2) - sqrt(3) / 4) - (1/2) * (arcsec(-7) - sqrt(48) / 49) =>

(1/2) * (arcsec(-2) - arcsec(-7) - sec(3)/4 + 4 * sqrt(3) / 49)

multiply top and bottom by x , let u² = 1- x² to get to [ - u² du ] / [ (1-u)² (1 + u)² ]

now do a partial fraction decomposition...- u² / [ ( 1 - u)² ( 1 + u )² ] =

a / (1-u) + b/(1-u)² + c/ (1 + u ) + d / (1 + u )²...find a,b,c,d and then

do the easy integrations