What is #int (x^3-2x^2+6x+9 ) / (2x^2- x +3 )#?

Question

What is #int (x^3-2x^2+6x+9     ) / (2x^2-   x  +3     )#?

Christopher Agresta · Accepted Answer

#\int (x^3-2x^2+6x+9)/(2x^2-x+3) "d"x = 1/4 (2x^2-3x) + 3/16 lnabs(2x^2-x+3) + 78/(16sqrt(23))arctan((4x+1)/(sqrt(23))) + C#.

#(x^3-2x^2+6x+9)/(2x^2-x+3)#. Since the quadratic on the denominator is irreducible, we can write this quotient in the form of a partial fraction decomposition. #Ax+B+(Cx+D)/(2x^2-x+3)#. When this fraction is simplified, it yields #(2Ax^3-Ax^2+3Ax+2Bx^2-Bx+3B+Cx+D)/(2x^2-x+3)#, #(2Ax^3+(2B-A)x^2+(C+3A-B)x+(D+3B))/(2x^2-x+3)#. Through coeffecient equations, #A=1/2#, #2B-1/2=-2 \implies B = -3/4# #C+3/2+3/4=6 \implies C = 15/4# #D+3B=9 \implies D = 45/4#. Then #(x^3-2x^2+6x+9)/(2x^2-x+3) = 1/4 (2x-3+(15x+45)/(2x^2-x+3))#. #\int (x^3-2x^2+6x+9)/(2x^2-x+3) "d"x = 1/4 \int 2x-3 "d"x + 3/4 \int (x+3)/(2x^2-x+3) "d"x# #\int (x^3-2x^2+6x+9)/(2x^2-x+3) "d"x = 1/4 (2x^2-3x) + 3/4 \int (x+3)/(2x^2-x+3) "d"x# We are then left with another integral to evaluate, #\int (x+3)/(2x^2-x+3) "d"x#. One significant general integral finding (particularly in the context of rational functions) is that, #\int (("f"'(x))/("f"(x))) "d"x = lnabs("f"(x))+C#. If we say #"f"(x) = 2x^2-x+3#, #"f"'(x)=4x-1#. If we make our numerator look like this, we will reduce the integral further. We may state, #\int (x+3)/(2x^2-x+3) "d"x = 1/4 \int (4x-1+13)/(2x^2-x+3) "d"x#, #\int (x+3)/(2x^2-x+3) "d"x = 1/4 \int (4x-1)/(2x^2-x+3) "d"x + 13/4 \int 1/(2x^2-x+3)#, #\int (x+3)/(2x^2-x+3) "d"x = 1/4 lnabs(2x^2-x+3) + 13/4 \int 1/(2x^2-x+3)#. After that, switching, #\int (x^3-2x^2+6x+9)/(2x^2-x+3) "d"x = 1/4 (2x^2-3x) + 3/16 lnabs(2x^2-x+3) + 39/16 \int 1/(2x^2-x+3) "d"x#. The last integral to evaluate is the arctangent integral, which can only be solved by completing the square because the quadratic on the denominator is irreducible (no real factors). #2x^2-x-3=2((x-1/4)^2 +23/16)#, #2x^2-x-3=2(x-1/4)^2+23/8#, #2x^3-x-3=1/8*(16(x-1/4)^2+23)#, #2x^3-x-3=1/8*((4x-1)^2+23)#. Then #int (1)/(2x^2-x+3) "d"x = \int (8)/((4x-1)^2+23) "d"x#. It is a typical outcome that, #\int 1/(x^2+a^2) = 1/a arctan(x/a) + C#. In our case, #a=sqrt(23)#. Applying the inverse chain rule to this outcome yields #\int (8)/((4x-1)^2+23) "d"x = 8 * (1/sqrt(23)*arctan((4x+1)/(sqrt(23))))/4 + C#, #\int (8)/((4x-1)^2+23) "d"x = 2/sqrt(23)*arctan((4x+1)/(sqrt(23))) + C#. At last, #\int (x^3-2x^2+6x+9)/(2x^2-x+3) "d"x = 1/4 (2x^2-3x) + 3/16 lnabs(2x^2-x+3) + 78/(16sqrt(23))arctan((4x+1)/(sqrt(23))) + C#.

Cameron Alexander · Answer

undefined To find the integral of $ \frac{x^3 - 2x^2 + 6x + 9}{2x^2 - x + 3} $, you would typically use the method of partial fraction decomposition followed by integration. However, since the partial fraction decomposition can be quite complex for this expression, it would be necessary to perform the division first to simplify the expression before attempting integration.