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

A Repeated Irreducible 2nd-degree Polynomial in The Denominator

Next, consider

Just as above, we can split x + 6 into (x − 4) + 10, and treat the part containing x − 4 via the substitution

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

This leaves us with

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:

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

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

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

and

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