u-substitution

f(u(x)) u'(x) dx = f(u) du

when to use: this is the most commonly used technique. It works when there is both an expression and the derivative of the expression within the integrant. It is also often useful if the integral is off by a constant to another integral that you know how to solve.

example:
(2x) sin(x²) dx

strategy: Here there is both the expression "x²" and its derivative "2x" inside the integrant. Therefore, we make a "u-substitution" by setting u=x², which is the original expression. By taking the derivative, we also find that du=2x dx.

solution:
   Let u = x²
   Then du = 2x dx
Substitute
  (2x) sin(x²) dx =  sin(u) du = -cos(u) + C.

Finally, we substitute back our expression for u:
   = COs(x²) + C

additional examples:

 (a+ bx) dx
1/(a + bx) dx
(a + bx)¹⁰ dx
x⁴ e^{(x^5)} dx
sin³(x) COs(x) dx

Integration by Parts

u dv = uv -  v du

when to use: when the above techniques do not work and one part of the integrant is easily integral while another part of the integrant is easily differentiable. The integrant is usually a product of two expressions.

Example:
x ln(x) dx
strategy: We set u equal to the part that is easily differentiable and dv equal to the part that is easily integratible (dv always includes the dx).  Integration by parts attempts to re-express the integral in terms of a new integral that contains the derivative of u and the antiderivative of dv.

"x" is easy to both differentiate and integrate. However, "ln(x)" is easy to differentiate but difficult to integrate. Therefore, we let u=ln(x) (the part that gets differentiated) and let dv=x dx (the part that gets integrated and includes the differential).

solution:
   Let u = ln(x) and dv = x dx
   Then du = (1/x) dx and v = x²/2.
u dv = UV -  v du = ln(x) (x²/2) -  (x²/2) (1/x) dx
The new integral can now be solved easily.

multiple applications
x^k e^x dx
special case
e^x sin(x) dx
e^x COs(x) dx

Special Trig Integrals
 
  secⁿ(x) tan^odd(x) dx =   sec^n-1(x) tan^even(x) [sec(x)tan(x)] dx
 

Trig Substitutions
when to use: when the above techniques do not work and there is a radical (square root) with an ugly expression inside it that you wish to get rid of.
Example:
 (1 + x²) dx
choose
   x = tan()
then
 (x² + 1) = (tan²() + 1) = (sec²()) = sec()
    dx = sec²() d
results in
  (x² + 1) dx =  (tan²() + 1) =  (sec²()) =  sec()

Partial Fractions
 
 

ax+b  (mx+n)(ox+p)

=

A  (Mx+n)

+

B  (ox+p)

when to use: when the integrant is a rational polynomial (and the degree of the numerator is less than that of the denominator) and the denominator is factorable.  The partial fraction technique splits the integral into multiple more manageable rational integrals.

 
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