1) To solve ∫[(x+1)/(2x²-4x+9)]dx we note that the solution has the form of ln(denominator) and arctan(?), so first differentiate the quadratic: 4x-4. To make the numerator the derivative of the denominator we note that ¼(4x-4)+2=x+1. Therefore we can rewrite the integral:
¼∫[(4x-4)/(2x²-4x+9)]dx+2∫dx/(2x²-4x+9).
The first part integrates nicely as ¼ln[a(2x²-4x+9)] where a is a constant of integration (combined from both integrals).
That leaves us with the other integral. We can write 2x²-4x+9 as 2(x-1)²+7.
We know the standard integral of 1/(x²+p²) is (1/p)arctan(x/p), so, using appropriate substitutions we can integrate:
2∫dx/(2(x-1)²+7)=∫dx/((x-1)²+(7/2)).
So x-1 replaces x and p=√(7/2), and the integral is:
√(2/7)arctan[√(2/7)(x-1)].
The full integration is:
¼ln[a(2x²-4x+9)]+√(2/7)arctan[√(2/7)(x-1)].
2) ∫[(1/x)/√(x²-1)]dx=∫dx/(x√(x²-1)).
Let x=secθ, dx=secθtanθdθ=x√(x²-1)dθ.
So the integral becomes ∫dθ=θ=arcsec(x)+c where c is the constant of integration.