Xojessilynox's question at Yahoo Answers involving integration

[Chris L T521]

Here is the question.

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.

 

[Chris L T521]

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.

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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
\[\begin{aligned} \int\frac{x+4}{(x+4)^2+1}\,dx \xrightarrow{u=(x+4)^2+1}{} \frac{1}{2}\int\frac{\,du}{u} &= \frac{1}{2}\ln|u|+C\\ &=\frac{1}{2}\ln|(x+4)^2+1|+C \\ &= \frac{1}{2}\ln(x^2+8x+17)+C.\end{aligned}\]

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Next, let us consider
\[\int\frac{4}{(x+4)^2+1}\,dx\]
Making the substitution $u=x+4\implies \,du=\,dx$, we see that
\[\begin{aligned} \int\frac{4}{(x+4)^2+1}\,dx \xrightarrow{u=x+4}{} \int\frac{4}{u^2+1}\,du &= 4\arctan(u)+C\\ &= 4\arctan(x+4)+C\end{aligned}\]

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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,
\[\begin{aligned} -\tfrac{3}{2}\cos^2\vartheta -\tfrac{13}{2}\vartheta- \tfrac{13}{2}\sin\vartheta\cos\vartheta +C &= -\tfrac{3}{2} \left(\frac{1}{x^2+8x+17}\right) -\tfrac{13}{2}\arctan(x+4) -\tfrac{13}{2}\left(\frac{x+4}{x^2+8x+17}\right) +C \\ &= -\frac{1}{2}\left[\frac{3}{x^2+8x+17} +13\arctan(x+4) +\frac{13x+52}{x^2+8x+17} \right] +C \\ &= -\frac{1}{2}\left[\frac{13x+55}{x^2+8x+17} + 13\arctan(x+4)\right]+C\end{aligned}\]

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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!