integral of (4/(x times the square root of( x^2-1)) + (1+x+x^3)/(1+x^2)dx
for integrating 1/√(x^2-1) use x = secθ
Then x√(x^2-1) = secθtanθ
dx = secθtanθ dθ
The integral of the first term then becomes
∫4 dθ = 4θ = 4 arcsec(x)
for integrating 1/(1+x^2) use x = tanθ
dx = sec^2θ dθ
Then, you have
∫x + 1/(x^2+1) dx which becomes
x^2/2 + ∫dθ = x^2/2 + arctan(x)
The whole integral then is
x^2/2 + 4 arcsec(x) + arctan(x) + C
or
x^2/2 + 4 arctan√(x^2-1) + arctan(x) + C