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Topics in Calculus
Fundamental theorem
Limits of functions
Continuity
Mean value theorem

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.

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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{du\over u}={1 \over a}\ln\left|u\right|+C = {1 \over a} \ln\left|ax+b\right|+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 x2 − 8x + 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}\ln\left|u\right|+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}\right)^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} \arctan(w)+C={10 \over 3} \arctan\left({x-4 \over 3}\right)+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} \arctan\left({x-4 \over 3}\right) + 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}\right)^2+1\right)^8}\,dx.

Then we can use a trigonometric substitution:

\tan\theta={x-4 \over 3},\,
\left({x-4 \over 3}\right)^2+1=\tan^2\theta+1=\sec^2\theta,\,
 d\tan\theta =\sec^2\theta\,d\theta={dx \over 3}.\,

Then the integral becomes

\int {30/9^{8} \over \sec^{16}\theta} \sec^2\theta \,d\theta ={30 \over 9^{8}}\int \cos^{14} \theta \, d\theta.

By repeated applications of the half-angle formula

\cos^2\theta={1 \over 2}( 1 + \cos(2\theta)),

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

\tan(\theta)={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) = √(x2 −8x + 25).

Therefore we have

\sin(\theta) = {\mathrm{opposite} \over \mathrm{hypotenuse}} = {x-4 \over \sqrt{x^2 - 8x + 25}},
\cos(\theta) = {\mathrm{adjacent} \over \mathrm{hypotenuse}} = {3 \over \sqrt{x^2 - 8x + 25}},

and

\sin(2\theta) = 2\sin(\theta)\cos(\theta) = {6(x-4) \over x^2 - 8x + 25}.

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