Exponential Integrals:

integral e^(ax)sin(bx) dx = e^(ax)(asin(bx) - bcos(bx))/(a^2+b^2) + C integral e^(ax)cos(bx) dx = e^(ax)(acos(bx) + bsin(bx))/(a^2+b^2) + C (Derived by applying by parts twice and solving for I)

integral x^n * e^x dx: use by parts n times (tabular method) integral e^x * x^n = e^x * (sum_{k=0}^{n} (-1)^k * n!/(n-k)! * x^(n-k)) + C

Logarithmic Integrals:

integral ln(x) dx = xln(x) - x + C integral (ln(x))^2 dx = x(ln(x))^2 - 2xln(x) + 2x + C integral x^n * ln(x) dx = x^(n+1)[ln(x)/(n+1) - 1/(n+1)^2] + C (for n != -1) integral ln(x)/x dx = (ln(x))^2/2 + C (substitution u = ln(x))

Special Exponential Forms:

integral e^(sqrt(x)) dx: Let t = sqrt(x), x = t^2, dx = 2t dt = 2integral te^t dt = 2e^t(t-1) + C = 2e^(sqrt(x))(sqrt(x)-1) + C

integral e^x * (x-1)/x^2 dx: Write (x-1)/x^2 = 1/x - 1/x^2. Note d/dx(1/x) = -1/x^2. So this is e^x[f(x) + f'(x)] with f = 1/x... wait: f(x) + f'(x) = 1/x + (-1/x^2) = 1/x - 1/x^2 = (x-1)/x^2. Yes! integral = e^x/x + C