How do you use partial fraction decomposition to decompose the fraction to integrate #(-x - 38)/(2x^2 + 9x - 5) #?

Question

Emily Ammerman · Accepted Answer

# int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c#

First we need to factorise the denominator. To do this we need to find factors of #-10# (coefficient of #x^2 xx# coefficient of constant#= 2 xx -5#) that add up tp 9 (the coefficient of #x=9#). So the factors we seek are #-1# and #10# # :. 2x^2+9x-5 = 2x^2 +10x - x -5 # # :. 2x^2+9x-5 =2x(x+5) - (x+5) # # :. 2x^2+9x-5 =(x+5)(2x-1) # Therefore we can write the integrand as follows: # (-x-38)/(2x^2+9x-5) = (-x-38)/((x+5)(2x-1))# And thge partial fraction decomposition will be: # \ \ \ \ \ (-x-38)/(2x^2+9x-5) = A/(x+5) + B/(2x-1)# # :. (-x-38)/(2x^2+9x-5) = (A(2x-1)+B(x+5))/((x+5)(2x-1)) # # :. \ \ \ \ (-x-38) = A(2x-1)+B(x+5) # Subs #x=-5 => -(-5)-38=A(-10-1) + 0# # :. -11A=-33 => A=3 # And Sub #x=1/2=>-1/2-38 = B(1/2+5) # # :. 11/2B=-77/2 => B=-7# Hence, the partial fraction decomposition of the integrand is: # (-x-38)/(2x^2+9x-5) = 3/(x+5) -7/(2x-1)# And so; # int(-x-38)/(2x^2+9x-5) dx= int 3/(x+5) -7/(2x-1) dx# And integrating we get: # int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c#

William Abdalla · Answer

undefined To decompose the fraction (-x - 38)/(2x^2 + 9x - 5) using partial fraction decomposition, follow these steps:

1. Factor the denominator polynomial: 2x^2 + 9x - 5.
2. Write the fraction as a sum of simpler fractions with unknown numerators: (-x - 38)/(2x^2 + 9x - 5) = A/(2x - 1) + B/(x + 5), where A and B are constants to be determined.
3. Multiply both sides of the equation by the denominator of the original fraction to clear the fractions.
4. Solve for the constants A and B by equating the numerators of the fractions obtained in step 3.
5. Once you have found the values of A and B, rewrite the original fraction as a sum of the partial fractions.
6. Finally, integrate each partial fraction separately.

The detailed calculations for finding the constants A and B can be done through various methods such as equating coefficients or using substitution.