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I want to integrate (2x+1)/((x^2 - 6x + 14)^3)

I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14 I find that it makes turning the top of the fraction into terms of u very difficult

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3 Answers 3

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$$\begin{align} \int \frac{2x + 1} {(x^2 - 6x + 14)^3} & = \int \frac{2x - 6 + 6 + 1}{(x^2 - 6x + 14)^3}\,dx \\ \\ & = \int\dfrac{2x - 6}{(x^2 - 6x + 14)^3}\,dx + \int \frac{7}{(x^2 - 6x + 14)^3}\,dx \\ \\ &= \int\dfrac{\overbrace{2x - 6}^{du}}{(\underbrace{x^2 - 6x + 14}_{u})^3}\,dx + \int \frac{7}{(\underbrace{x^2 - 6x + 9}_{(x - 3)^2} + \underbrace{5}_{(\sqrt 5)^2})^3}\,dx \end{align}$$

The left-most integral: using $u = x^2 - 6x + 14 \implies du = 2x - 6, \,dx$ so we integrate $\displaystyle \int \frac {du}{u^3}$.

The right-most integral, we can use the substitution $x - 3 = \sqrt 5\tan \theta \implies dx = \sqrt 5 \sec^2 \theta\,d\theta$, to obtain the integral $$\displaystyle 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \tan^2 \theta + 5)^3}= 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 (\tan^2 \theta +1))^3} = 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \sec^2\theta)^3} = \cdots$$

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are you suggesting something other than Trigonometric substitution for the last half? –  lab bhattacharjee Nov 27 '13 at 15:36
    
@amWhy: Needs a TU -> +1 –  Amzoti Nov 28 '13 at 2:15

Using Trigonometric substitution

put $x-3=\sqrt5\tan\theta$ as $\displaystyle x^2-6x+14=(x-3)^2+(\sqrt5)^2$

$$\int \frac{2x+1}{(x^2-6x+14)^3}dx=\int\frac{2(\sqrt5\tan\theta+3)+1}{(5\sec^2\theta)^3}\sqrt5\sec^2\theta d\theta$$

$$=\frac1{25\sqrt5}\int(2\sqrt5\tan\theta+7)\cos^4\theta d\theta=\cdots$$

$$=\frac2{25}\int\sin\theta\cos^3\theta d\theta+\frac7{25\sqrt5}\int\cos^4\theta d\theta$$

The first integral can be managed by substituting $\cos\theta =u$

For the second one use Double-Angle Formulas , $\displaystyle \cos^4\theta=\frac{(2\cos^2\theta)^2}4=\frac{(1+\cos2\theta)^2}4=\frac14+\frac{\cos2\theta}2+\frac{\cos^22\theta}4$

Again, $\displaystyle2\cos^22\theta=1+\cos4\theta$

As the principal value $\arctan$ lies $\left[-\frac\pi2,\frac\pi2\right]$

$\displaystyle\cos\theta=+\frac1{\sqrt{1+\tan^2\theta}}=\frac{\sqrt5}{\sqrt{x^2-6x+14}}$

and $\displaystyle\sin\theta=\tan\theta\cdot\cos\theta$

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I'd suggest using $y = x^2 - 6x + 14$. Then $du = 2x - 6$, which is almost what you've got in the numerator. So split your fraction:

$$ \frac{2x + 1} {(x^2 - 6x + 14)^3} = \frac{2x - 6} {(x^2 - 6x + 14)^3} + \frac{7} {(x^2 - 6x + 14)^3} $$

That splits your integral into two parts; the first part can be integrated with the substitution I suggested. The second part will require a bit more work, completing the square in the denominator and doing a tangent-substitution, probably.

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and the substitution you want to use in that second integral is the one suggested by lab bhattacharjee above. –  John Nov 27 '13 at 15:22
    
Thank you I think i've completed it now! –  Lauren Nov 27 '13 at 15:53

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