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In integral calculus, the use of partial fractions is required to integrate the general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of partial fractions. Each partial fraction has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a function. If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.
For an account of how to find this partial fraction expansion of a rational function, see partial fraction.
This article is about what to do ''after'' finding the partial fraction expansion, when one is trying to find the function's antiderivative.

Contents

A 1st-degree polynomial in the denominator

A repeated 1st-degree polynomial in the denominator

An irreducible 2nd-degree polynomial in the denominator

A repeated irreducible 2nd-degree polynomial in the denominator

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A 1st-degree polynomial in the denominator

The substitution ''u'' = ''ax'' + ''b'', ''du'' = ''a'' ''dx'' reduces the integral
:$int\; \{1\; over\; ax+b\},dx$
to
:$int\; \{1\; over\; u\},\{du\; over\; a\}=\{1\; over\; a\}int\{duover\; u\}=\{1\; over\; a\}lnleft|u$
ight|+C = {1 over a} lnleft|ax+b
ight|+C.

A repeated 1st-degree polynomial in the denominator

The same substitution reduces such integrals as
:$int\; \{1\; over\; (ax+b)^8\},dx$
to
:$int\; \{1\; over\; u^8\},\{du\; over\; a\}=\{1\; over\; a\}int\; u^\{-8\},du\; =\; \{1\; over\; a\}\; cdot\{u^\{-7\}\; over(-7)\}+C\; =\; \{-1\; over\; 7au^7\}+C\; =\; \{-1\; over\; 7a(ax+b)^7\}+C.$

An irreducible 2nd-degree polynomial in the denominator

Next we consider such integrals as
:$int\; \{x+6\; over\; x^2-8x+25\},dx.$
The quickest way to see that the denominator ''x''² − 8''x'' + 25 is irreducible is to observe that its discriminant is negative. Alternatively, we can complete the square:
:$x^2-8x+25=(x^2-8x+16)+9=(x-4)^2+9,$
and observe that this sum of two squares can never be 0 while ''x'' is a real number.
In order to make use of the substitution
:$u=x^2-8x+25,$
:$du=(2x-8),dx$
:$du/2=(x-4),dx$
we would need to find ''x'' − 4 in the numerator. So we decompose the numerator ''x'' + 6 as (''x'' − 4) + 10, and we write the integral as
:$int\; \{x-4\; over\; x^2-8x+25\},dx\; +\; int\; \{10\; over\; x^2-8x+25\},dx.$
The substitution handles the first summand, thus:
:$int\; \{x-4\; over\; x^2-8x+25\},dx\; =\; int\; \{du/2\; over\; u\}$
= {1 over 2}lnleft|u
ight|+C
= {1 over 2}ln(x^2-8x+25)+C.
Note that the reason we can discard the absolute value sign is that, as we observed earlier, (''x'' − 4)² + 9 can never be negative.
Next we must treat the integral
:$int\; \{10\; over\; x^2-8x+25\}\; ,\; dx.$
First, complete the square, then do a bit more algebra:
:$int\; \{10\; over\; x^2-8x+25\}\; ,\; dx$
= int {10 over (x-4)^2+9} , dx
= int {10/9 over left({x-4 over 3}
ight)^2+1},dx
Now the substitution
:$w=(x-4)/3,$
:$dw=dx/3,$
gives us
:$\{10\; over\; 3\}int\; \{dw\; over\; w^2+1\}$
= {10 over 3} rctan(w)+C={10 over 3} rctanleft({x-4 over 3}
ight)+C.
Putting it all together,
:$int\; \{x\; +\; 6\; over\; x^2-8x+25\},dx$
= {1 over 2}ln(x^2-8x+25) + {10 over 3} rctanleft({x-4 over 3}
ight) + C.

A repeated irreducible 2nd-degree polynomial in the denominator

Next, consider
:$int\; \{x+6\; over\; (x^2-8x+25)^\{8\}\},dx.$
Just as above, we can split ''x'' + 6 into (''x'' − 4) + 10, and treat the part containing ''x'' − 4 via the substitution
:$u=x^2-8x+25,,$
:$du=(2x-8),dx$
:$du/2=(x-4),dx.$
This leaves us with
:$int\; \{10\; over\; (x^2-8x+25)^\{8\}\},dx.$
As before, we first complete the square and then do a bit of algebraic massaging, to get
:$int\; \{10\; over\; (x^2-8x+25)^\{8\}\},dx$
=int {10 over ((x-4)^2+9)^{8}},dx
=int {10/9^{8} over left(left({x-4 over 3}
ight)^2+1
ight)^8},dx.
Then we can use a trigonometric substitution:
:$an\; heta=\{x-4\; over\; 3\},,$
:$left(\{x-4\; over\; 3\}$
ight)^2+1= an^2 heta+1=sec^2 heta,,
:$d\; an\; heta=sec^2\; heta,d\; heta=\{dx\; over\; 3\}.,$
Then the integral becomes
:$int\; \{30/9^\{8\}\; over\; sec^\{16\}\; heta\}\; sec^2\; heta\; ,d\; heta$
={30 over 9^{8}}int cos^{14} heta , d heta
By repeated applications of the half-angle formula
:$cos^2\; heta=\{1\; over\; 2\}+\{1\; over\; 2\}\; cos(2\; heta),$
one can reduce this to an integral involving no higher powers of cos θ higher than the 1st power.
Then one faces the problem of expression sin(θ) and cos(θ) as functions of ''x''. Recall that
:$an(\; heta)=\{x\; -\; 4\; over\; 3\},$
and that tangent = opposite/adjacent. If the "opposite" side has length ''x'' − 4 and the "adjacent" side has length 3, then the Pythagorean theorem tells us that the hypotenuse has length √((''x'' − 4)² + 3²) = √(''x''² −8''x'' + 25).
Therefore we have
:$sin(\; heta)\; =\; \{mathrm\{opposite\}\; over\; mathrm\{hypotenuse\}\}\; =\; \{x-4\; over\; sqrt\{x^2\; -\; 8x\; +\; 25\}\},$
:$cos(\; heta)\; =\; \{mathrm\{adjacent\}\; over\; mathrm\{hypotenuse\}\}\; =\; \{3\; over\; sqrt\{x^2\; -\; 8x\; +\; 25\}\},$
and
:$sin(2\; heta)\; =\; 2sin(\; heta)cos(\; heta)\; =\; \{6(x-4)\; over\; x^2\; -\; 8x\; +\; 25\}.$

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★ Partial Fraction Expander