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
This leaves us with
As before, we first complete the square and then do a bit of algebraic massaging, to get
Then we can use a trigonometric substitution:
Then the integral becomes
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
Read more about this topic: Partial Fractions In Integration
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