This is a standard irreducible quadratic factor. ∫ dx/(x^2+1) = tan⁻¹(x) + C.
Standard form: ∫ dx/(x^2 + a^2) = (1/a) tan⁻¹(x/a) + C. Here a = 2.
Let u = x^2+1, then du = 2x dx. So ∫ x/(x^2+1) dx = (1/2) ln(x^2+1) + C.
Reducible quadratic factor. Partial fractions: 1/((x-1)(x-2)) = 1/(x-1) - 1/(x-2). Integration gives ln|(x-1)/(x-2)| + C.
Reducible quadratic factor. Partial fractions: 1/(x^2-1) = (1/2)(1/(x-1) - 1/(x+1)).
Complete the square: x^2+2x+5 = (x+1)^2+4. Then ∫ dx/((x+1)^2+2^2) = (1/2) tan⁻¹((x+1)/2) + C.
Denominator derivative (2x+3) appears in numerator. So ∫ (2x+3)/(x^2+3x+2) dx = ln|x^2+3x+2| + C.
Irreducible × linear → partial fractions. After decomposition, result is (1/2) ln|(x-1)^2/(x^2+1)| + tan⁻¹(x) + C.
Let u = x^2+4, then du = 2x dx. So ∫ x/(x^2+4) dx = (1/2) ln(x^2+4) + C.
Standard irreducible quadratic: ∫ dx/(u^2+a^2) = (1/a) tan⁻¹(u/a). Here u = x+1, a = 3.