Integration with partial fractions

Question

Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember the constant of integration.)

∫ dx/(x√x-1)

Answer 1

Let u = √(x-1)

u² = x-1

x = u² + 1

dx = 2udu

∫ (2udu)/[((u²+1)u] = 2∫du/(u²+1) = 2tan^-1(u) = 2tan^-1(√(x-1))+C

desmos.com/calculator/nd9ilu6w19

Answer 2

I would make the substitution y²=x-1 so that x=y²+1 and dx=2y² dy

Then the integration can be performed by a simple u-substitution.

Answer 3

x = t²; dx = 2tdt; I = ∫dx/(x√x - 1) = ∫2tdt/(t³ - 1) = ∫2tdt/(t - 1)(t² + t + 1);

2t/(t - 1)(t² + t + 1) = a/(t - 1) + (bt + c)/(t² + t + 1); 2t = a(t² + t + 1) + (bt + c)(t - 1); a + b = 0; a - b + c = 2; a - c = 0; c = a; b = - a; 3a = 2; a = 2/3; b = - 2/3, c = 2/3

I = ∫(2/3)dt/(t - 1) - 2/3∫(t - 1)dt/(t² + t + 1) = 2/3lnIt - 1I - 1/3∫(2t +1 - 3)dt/(t² + t + 1) = 2/3lnIt - 1I - 1/3ln(t² + t + 1) + ∫dt/[(t + 1/2)² + 3/4] = 2/3lnI√x -1I - 1/3ln(x + √x + 1) + 2/√3tan^-1[( √x +1 /2)2/√3] + C