# How do I integrate a Fraction ?

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There are two types of fraction integrals you might come across. Fractions where the denominator is factorisable and fractions where it is not. So first try and factorise the denominator.

Type 1 Factorisable denominator

∫ 1/ x+3x + 2 dx = ∫ 1/(x+2)(x+1) dx

You must split it into two fractions using the method of partial fractions.

Let A/(x+2) + B/(x+1) = 1/(x+2)(x+1) and by adding the fractions you can solve for A = -1 , B = 1 . You then need only integrate using the rule for the derivative of ln x .

∫ 1/ (x+1) - 1/(x+2) dx = ln (x+1) - ln (x+2) + c

Type 2 Denominator unfactorisable

∫ 1/ x+2x + 9 dx Instead of factorising, complete the square of the denominator.

∫ 1/(x+1)2 + 8 dx Now you must choose a suitable substitution to turn the denominator into one term. I would choose tan2x + 1 + sec2x

So let x+1 = 8 tan y

Now our integral becomes  ∫ 1/ 8 sec2 y dx

Also multiply by the derivative of your substitution so you can integrate in terms of y for example ∫ f(x) dx = ∫ f(y) dx/dy dy because the dy's cancel.

dx/dy = 8 sec2y

∫ 8 sec2y / 8 sec2 y dy = ∫ 1/8 dy = 1/8 y + c

Now you must use your substitution get the solution in terms of x.

tan-1 (x+1/8) 8 + c