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How do you integrate #int(x+1)/((6x^2+4)(x-5))# using partial fractions?

Answer 1

Scarlett Allison

See below.

I used basic fractions method.

Inıtıially, I decomposed integrand into basic fractions,

#(x+1)/[(6x^2+4)*(x-5)]=A/(x-5)+(Bx+C)/(6x^2+4)#

After expanding denominator,

#A*(6x^2+4)+(Bx+C)*(x-5)=x+1#

#(6A+B)*x^2+(C-5B)*x+4A-5C=x+1#

After equating coefficients, I found

#6A+B=0#, #C-5B=1# and #4A-5C=1# equations.

After solving them, #A=3/77, B=-18/77 and C=-13/77#

Thus,

#int ((x+1)dx)/[(6x^2+4)*(x-5)]#

=#3/77##int dx/(x-5)#-#1/77##int ((18x+13)*dx)/(6x^2+4)#

=#3/77*ln(x-5)#-#3/154##int (12x*dx)/(6x^2+4)#-#13/77##int dx/(6x^2+4)#

=#3/77*ln(x-5)-3/154*ln(6x^2+4)-(13sqrt(6))/924*arctan((3x)/sqrt(6))+C#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.