Let u = x^2 + 1, then du = 2x dx. Hence ∫ (2x)/(x^2+1) dx = ∫ du/u = ln(x^2+1) + C.

Let u = 3x, then du = 3 dx → dx = du/3. Hence ∫ cos(3x) dx = (1/3) sin(3x) + C.

Let u = 2x, then du = 2 dx → dx = du/2. Hence ∫ e^(2x) dx = (1/2) e^(2x) + C.

Let u = sin(x), then du = cos(x) dx. Hence ∫ u^2 du = u^3/3 + C = (sin^3(x))/3 + C.

This is a standard substitution integral. ∫ dx / √(1 - x^2) = sin⁻¹(x) + C.

Let u = x^4 + 1, then du = 4x^3 dx → dx = du/(4x^3). Hence ∫ x^3/(x^4+1) dx = (1/4) ln(x^4+1) + C.

Let u = x^2, then du = 2x dx. So ∫ 2x e^(x^2) dx = ∫ e^u du = e^u + C = e^(x^2) + C.

Let u = ln(x), then du = dx/x. Hence ∫ (1/(x ln(x))) dx = ∫ du/u = ln|ln(x)| + C.

Since tan(x) = sin(x)/cos(x), let u = cos(x), du = -sin(x) dx. Hence ∫ tan(x) dx = -ln|cos(x)| + C.

Let u = 5x, then du = 5 dx → dx = du/5. Hence ∫ sec^2(5x) dx = (1/5) tan(5x) + C.