Try letting dv be the most complicated portion of the integrand that fits a basic integration rule. Then u will be the remaining factor(s) of the integrand. Try letting u be the portion of the integrand whose derivative is a function simpler than u. Then dbv will be the remaining factor(s) of the integrand. Note that dv always includes the dx of the original integrand.
1. Factor the denominator 2. Break up the fraction on the right into a sum of fractions, where each factor of the denominator in Step 1 becomes the denominator of a separate fraction. Then put unknowns in the numerator of each fraction. 3. Multiply both sides of this equation by the denominator of the left side. 4. Take the roots of the linear factors and plug them –one at a time-into x in the equation from Step 3, and solve for unknowns. 5. Plug these results into the A and B in the equation from Step 2. 6. Split up the original integral into the partial fractions from Step 5 and you’re home free!
1. Factor the denominator 2. Break up the fraction into a sum of “partial fractions” 3. Multiply both sides of this equation by the left-side denominator. 4. Take the roots of the linear factors and plug them- one at a time- into x in the equation from Step 3, and then solve. 5. Plug into Step 3 equation the known values of A and B and any two values for x not used in Step 4 to get a system of two equations in C and D. 6. Solve the system of equations. 7. Split up the original integral and integrate.