MHB Xojessilynox's question at Yahoo Answers involving integration

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To evaluate the integral of the function (x^3 + 8x^2 + 26x - 3) / (x^2 + 8x + 17)^2, partial fraction decomposition is used, resulting in two simpler integrals. The first integral involves the logarithmic function, while the second requires trigonometric substitution. After performing the necessary substitutions and integrations, the final result combines logarithmic and arctangent terms. The complete evaluation yields a concise expression involving these functions, providing a comprehensive solution to the integration problem. The detailed steps ensure clarity in the integration process.
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Here is the question.

CALC 2: evaluate the integral x^3+8x^2+26x-3 / (x^2+8x+17)^2? said:
I can't find the answer to this question. I keep getting this wrong too just like all the other problems. I need help please and thank you!

Here is a link to the question:

CALC 2: evaluate the integral x^3+8x^2+26x-3 / (x^2+8x+17)^2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Re: xojessilynox's question at yahoo answers involving integration

Hi xojessilynox,

To evaluate $\displaystyle\int \frac{x^3+8x^2+26x-3}{(x^2+8x+17)^2}\,dx$, we proceed first by partial fractions.

The denominator contains a power of an irreducible quadratic, so our partial fraction decomposition takes on the form
\[\frac{x^3+8x^2+26x-3}{(x^2+8x+17)^2} = \frac{Ax+B}{x^2+8x+17} + \frac{Cx+D}{(x^2+8x+17)^2}.\]

Multiplying both sides by the common denominator gives us
\[\begin{aligned} x^3+8x^2+26x-3 &= (Ax+B)(x^2+8x+17)+Cx+D\\ &= Ax^3+(8A+B)x^2+(17A+8B+C)x+17B+D. \end{aligned}\]
Comparing coefficients gives rise to the following system of equations:
\[\left\{\begin{aligned} A &= 1\\ 8A+B &= 8\\ 17A+8B+C &= 26\\ 17B+D &= -3\end{aligned}\right.\]
Which has the solutions $A=1$, $B=0$, $C=9$ and $D=-3$ (Verify). Therefore,
\[\frac{x^3+8x^2+26x-3}{(x^2+8x+17)^2} = \frac{x}{x^2+8x+17} + \frac{9x-3}{(x^2+8x+17)^2}\]
and thus
\[\begin{aligned} \int\frac{x^3+8x^2+26x-3}{(x^2+8x+17)^2}\,dx &= \int\frac{x}{x^2+8x+17}\,dx + \int\frac{9x-3}{(x^2+8x+17)^2}\,dx\\ &= \int\frac{x}{(x+4)^2+1}\,dx + 3\int\frac{3x-1}{((x+4)^2+1)^2}\,dx\\ &= \int\frac{x+4}{(x+4)^2+1}\,dx - \int\frac{4}{(x+4)^2+1}\,dx +3\int\frac{3x-1}{((x+4)^2+1)^2}\,dx. \end{aligned}\]
Let's focus on each of these three integrals one at a time.


Consider
\[\int\frac{x+4}{(x+4)^2+1}\,dx.\]
Making the substitution $u=(x+4)^2+1\implies \,du=2(x+4)\,dx \implies \frac{1}{2}\,du=(x+4)\,dx$, we see that
\[\begin{aligned} \int\frac{x+4}{(x+4)^2+1}\,dx \xrightarrow{u=(x+4)^2+1}{} \frac{1}{2}\int\frac{\,du}{u} &= \frac{1}{2}\ln|u|+C\\ &=\frac{1}{2}\ln|(x+4)^2+1|+C \\ &= \frac{1}{2}\ln(x^2+8x+17)+C.\end{aligned}\]
Next, let us consider
\[\int\frac{4}{(x+4)^2+1}\,dx\]
Making the substitution $u=x+4\implies \,du=\,dx$, we see that
\[\begin{aligned} \int\frac{4}{(x+4)^2+1}\,dx \xrightarrow{u=x+4}{} \int\frac{4}{u^2+1}\,du &= 4\arctan(u)+C\\ &= 4\arctan(x+4)+C\end{aligned}\]
Finally, we consider
\[\int\frac{3x-1}{((x+4)^2+1)^2}\,dx\]
To compute this guy, we use the trig substitution $x+4=\tan\vartheta \implies x=\tan\vartheta-4$. Thus, $\,dx=\sec^2\vartheta\,d\vartheta$ and we now see that
\[\begin{aligned} \int\frac{3x-1}{((x+4)^2+1)^2}\,dx \xrightarrow{x+4=\tan\vartheta}{} \int\frac{3(\tan\vartheta -4)-1}{(\tan^2\vartheta+1)^2} \sec^2\vartheta\,d\vartheta &= \int\frac{3\tan\vartheta - 13}{(\sec^2\vartheta)^2} \sec^2\vartheta\,d\vartheta\\ &= \int\frac{3\tan\vartheta -13}{\sec^2\vartheta} \,d\vartheta\\ &= \int (3\tan\vartheta\cos^2\vartheta - 13\cos^2\vartheta) \,d\vartheta\\ &= \int (3\sin\vartheta\cos\vartheta -13\cos^2\vartheta) \,d\vartheta\\ &=\int (\tfrac{3}{2}\sin(2\vartheta)- \tfrac{13}{2}(1+\cos(2\vartheta))) \,d\vartheta\\ &= \int (\tfrac{3}{2}\sin(2\vartheta) -\tfrac{13}{2} -\tfrac{13}{2}\cos(2\vartheta)) \,d\vartheta\\ &= -\tfrac{3}{4}\cos(2\vartheta) -\tfrac{13}{2}\vartheta -\tfrac{13}{4}\sin(2\theta)+C\\ &= -\tfrac{3}{4}(2\cos^2\vartheta - 1) - \tfrac{13}{2}\vartheta -\tfrac{13}{4}(2\sin\vartheta\cos\vartheta) + C\\ &=-\tfrac{3}{2}\cos^2\vartheta +\tfrac{3}{4} -\tfrac{13}{2}\vartheta -\tfrac{13}{2}\sin\vartheta\cos\vartheta+C\\ &= -\tfrac{3}{2}\cos^2\vartheta -\tfrac{13}{2}\vartheta -\tfrac{13}{2}\sin\vartheta\cos\vartheta + C.\end{aligned}\]

Note that in the last line above, $\tfrac{3}{4}$ is a constant, so it can get absorbed in the integration constant $C$.

Now, to convert everything back into terms of $x$, we recall that $x+4=\tan\vartheta$. Constructing a right triangle with the side opposite of $\vartheta$ having length $x+4$ and adjacent side with length $1$, it follows that $\sin\vartheta=\dfrac{x+4}{\sqrt{(x+4)^2+1}} = \dfrac{x+4}{\sqrt{x^2+8x+17}}$, $\cos\vartheta=\dfrac{1}{\sqrt{x^2+8x+17}}$ and $\vartheta=\arctan(x+4)$. Therefore,
\[\begin{aligned} -\tfrac{3}{2}\cos^2\vartheta -\tfrac{13}{2}\vartheta- \tfrac{13}{2}\sin\vartheta\cos\vartheta +C &= -\tfrac{3}{2} \left(\frac{1}{x^2+8x+17}\right) -\tfrac{13}{2}\arctan(x+4) -\tfrac{13}{2}\left(\frac{x+4}{x^2+8x+17}\right) +C \\ &= -\frac{1}{2}\left[\frac{3}{x^2+8x+17} +13\arctan(x+4) +\frac{13x+52}{x^2+8x+17} \right] +C \\ &= -\frac{1}{2}\left[\frac{13x+55}{x^2+8x+17} + 13\arctan(x+4)\right]+C\end{aligned}\]


Putting it all together, we see that

\[\begin{aligned} \int\frac{x^3+8x^2+26x-3}{(x^2+8x+17)^2}\,dx &= \int\frac{x+4}{(x+4)^2+1}\,dx - \int\frac{4}{(x+4)^2+1}\,dx +3\int\frac{3x-1}{((x+4)^2+1)^2}\,dx \\ &= \left(\tfrac{1}{2}\ln(x^2+8x+17)\right) - \left(4\arctan(x+4)\right) + 3\left(-\frac{1}{2}\left[\frac{13x+55}{x^2+8x+17} + 13\arctan(x+4)\right]\right) + C \\ &= \frac{1}{2}\left[\ln(x^2+8x+17) - 8\arctan(x+4) -\frac{3(13x+55)}{x^2+8x+17} - 39\arctan(x+4)\right] + C\\ &= \boxed{\dfrac{1}{2} \left[\ln(x^2+8x+17) - 47\arctan(x+4) - \dfrac{3(13x+55)}{x^2+8x+17} \right] + C}\end{aligned}\]

I hope everything makes sense!
 
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