Partial Fractions in Integration - An Irreducible 2nd-degree Polynomial in The Denominator

An Irreducible 2nd-degree Polynomial in The Denominator

Next we consider such integrals as

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:

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

\begin{align} u & = x^2-8x+25 \\ du & =(2x-8)\,dx \\ du/2 & = (x-4)\,dx \end{align}

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

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

First, complete the square, then do a bit more algebra:

\begin{align} & {} \quad \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 = {10 \over 3} \int {1 \over \left({x-4 \over 3}\right)^2+1}\, \left({dx \over 3}\right) \end{align}

Now the substitution

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.

Read more about this topic:  Partial Fractions In Integration

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