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

Question

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

Emma Aldridge · Accepted Answer

#-1/2x^2-x+3/4ln(2x^2-x+3)+sqrt(23)/2arctan((4x-1)/sqrt(23))# #int(-2x^3-x^2+x+2)/(2x^2-x+3)dx# #=int(-2x^3+x^2-3x)/(2x^2-x+3)+(-2x^2+4x+2)/(2x^2-x+3)dx# (separate numerator) #=int-x+(-2x^2+x-3)/(2x^2-x+3)+(3x+5)/(2x^2-x+3)dx# (factor out #2x^2-x+3# from 1st fraction and separate second fraction so the numerator will divide evenly by the denominator) #=-1/2x^2+int-1+(3x-3/4)/(2x^2-x+3)+(23/4)/(2x^2-x+3)dx# (integrate #-x# with reverse power rule, factor out #2x^2-x+3# from 1st fraction, and separate second fraction. you don't need to write +C yet because another +C will be created from integrating the rest of the expression) #=-1/2x^2-x+int3/4((4x-1)/(2x^2-x+3))+(23/4)/((sqrt(2)x-sqrt(2)/4)^2+23/8)dx# (integrate #-1# and rewrite the fractions so it will be easier to integrate to #ln(f(x))# and #arctan(f(x))#) #=-1/2x^2-x+3/4ln|2x^2-x+3|+int(23/4)/((sqrt(2)x-sqrt(2)/4)^2+23/8)dx# (integrate the first fraction using #int(f'(x))/f(x)dx=ln(f(x))# #=-1/2x^2-x+3/4ln(2x^2-x+3)+23/4*1/sqrt(2)intsqrt(2)/((sqrt(2)x-sqrt(2)/4)^2+(sqrt(23/8))^2)dx# (change coefficients to prepare for integration to #arctan(f(x))# and change absolute value to parentheses because #2x^2-x+3# is always positive)# #=-1/2x^2-x+3/4ln(2x^2-x+3)+(23/4)/(sqrt(2*23/8))arctan((sqrt(2)x-sqrt(2)/4)/(sqrt(23/8)))+C# (integrate the fraction using #int(f'(x))/((f(x))^2+a^2)dx=1/aarctan(f(x)/a)+C# where a is a constant) #=-1/2x^2-x+3/4ln(2x^2-x+3)+(23/4)/(sqrt(23/4))arctan((sqrt(2*8)x-sqrt(2*8)/4)/(sqrt(23)))# (simplify) #=-1/2x^2-x+3/4ln(2x^2-x+3)+sqrt(23/4)arctan((4x-1)/sqrt(23))# #=-1/2x^2-x+3/4ln(2x^2-x+3)+sqrt(23)/2arctan((4x-1)/sqrt(23))#

Emily Ammerman · Answer

undefined To integrate $ \frac{-2x^3 - x^2 + x + 2}{2x^2 - x + 3} $, we can perform polynomial long division or use partial fraction decomposition.

Using polynomial long division, divide $ -2x^3 - x^2 + x + 2 $ by $ 2x^2 - x + 3 $, which yields a quotient and a remainder.

Once the division is done, the integral will be expressed as the integral of the quotient plus the integral of the remainder over the divisor.

Alternatively, using partial fraction decomposition, we express the rational function $ \frac{-2x^3 - x^2 + x + 2}{2x^2 - x + 3} $ as the sum of simpler fractions.

We start by factoring the denominator $ 2x^2 - x + 3 $ if possible. If not, we can use the quadratic formula.

Then, we express the rational function as:

$$ \frac{-2x^3 - x^2 + x + 2}{2x^2 - x + 3} = \frac{Ax + B}{2x^2 - x + 3} $$

We then solve for constants $ A $ and $ B $ by equating coefficients and find their values.

After decomposing the rational function into partial fractions, we integrate each term separately.

This process will yield the integral of the original rational function.