How do you integrate #int(x+1)/((x^2-2)(3x-5))# using partial fractions?

Answer 1

#int(x+1)/((x^2-2)(3x-5))=int(-8/7x-11/7)/(x^2-2)+int(24/7)/(3x-5)->-8/7intx/(x^2-2)-11/7int1/(x^2-2)+24/7int1/(3x-5)->-4/7ln(x^2-2)-(11sqrt2)/28 ln((x-sqrt2)/(x+sqrt2))+8/7ln(3x-5)#

To integrate we first have to resolve into partial fraction: #(x+1)/((x^2-2)(3x-5))=(Ax+B)/((x^2-2))+C/(3x-5)#

#x+1=(Ax+B)(3x-5)+C(x^2-2)#

#x+1=3Ax^2+3Bx-5Ax-5B+Cx^2-2C#

#0=3A+C,3B-5A=1,-5B-2C=1#

#A=8/-7,B=-11/7,C=24/7#

#(x+1)/((x^2-2)(3x-5))=(8/-7x+(-11)/7)/((x^2-2))+(24/7)/(3x-5)# Then integrate each part.

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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.

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