can you help with this math problem? I dont know how to start? There is any good method? $$\int\frac{x+1}{(x^2-x-5)^3} dx$$
Thank you
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Very similar question ($\int \frac{x+1}{\left(x^2-x+8\right)^3}, dx$): http://math.stackexchange.com/questions/1553924/evaluate-int-fracx1x2-x83-dx – user236182 Dec 02 '15 at 19:48
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The difference here is that $x^2-x-5$ has real roots, so you can solve it using partial fraction decompositions. – user236182 Dec 02 '15 at 19:50
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@user236182: Sorry to ask this here but how one usually finds similar questions or duplicates by searching? How did you do this in case of this question? :) – Hosein Rahnama Dec 03 '15 at 20:19
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@H.R. Well, it's not an exact duplicate, and the question was asked the day before, and I answered it myself. – user236182 Dec 03 '15 at 20:27
3 Answers
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$$\int \frac{x+1}{\left(x^2-x-5\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x-5\right)}{\left(x^2-x-5\right)^3}+\int\frac{3}{\left(x^2-x-5\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x-5\right)^2}+96\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}\right)$$
Let $2x-1=u$. We're searching for:
$$\int \frac{1}{(u+\sqrt{21})^3(u-\sqrt{21})^3}\, du$$
Use partial fraction decomposition. After some work you'll find that (WolframAlpha gives it):
$$\frac{1}{(u+\sqrt{21})^3(u-\sqrt{21})^3}=\frac{-1}{2352 \sqrt{21} (u+\sqrt{21})}+\frac{-1}{2352 (u+\sqrt{21})^2}$$
$$+\frac{-1}{168 \sqrt{21} (u+\sqrt{21})^3}+\frac{-1}{2352 \sqrt{21} (\sqrt{21}-u)}$$
$$+\frac{-1}{2352 (\sqrt{21}-u)^2}+\frac{-1}{168 \sqrt{21} (\sqrt{21}-u)^3}$$
$$\int \frac{1}{(u+\sqrt{21})^3(u-\sqrt{21})^3}\, du=\frac{-1}{2352\sqrt{21}}\int \frac{d(u+\sqrt{21})}{u+\sqrt{21}}$$
$$+\frac{-1}{2352}\int \frac{d(u+\sqrt{21})}{(u+\sqrt{21})^2}+\frac{-1}{168 \sqrt{21}}\int\frac{d(u+\sqrt{21})}{(u+\sqrt{21})^3}$$
$$+\frac{1}{2352 \sqrt{21}} \int\frac{d(\sqrt{21}-u)}{\sqrt{21}-u}+$$
$$+\frac{1}{2352}\int\frac{d(\sqrt{21}-u)}{(\sqrt{21}-u)^2}+\frac{1}{168 \sqrt{21}}\int\frac{d(\sqrt{21}-u)}{(\sqrt{21}-u)^3},$$
which is equal to:
$$\frac{-1}{2352\sqrt{21}}\ln|u+\sqrt{21}|+\frac{1}{2352(u+\sqrt{21})}+\frac{1}{336 \sqrt{21}(u+\sqrt{21})^2}$$
$$+\frac{1}{2352 \sqrt{21}}\ln|\sqrt{21}-u|+\frac{-1}{2352(\sqrt{21}-u)}+\frac{-1}{336 \sqrt{21}(\sqrt{21}-u)^2}$$
user236182
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Hint
1) Use partial fraction to decompose the integrand
$\eqalign{ & f = {{x + 1} \over {{{\left( {{x^2} - x - 5} \right)}^3}}} \cr & \,\,\,\, = \mathop \sum \limits_{r = RootOf\left( {{x^2} - x - 5} \right)} {{{r \over {147}} + {1 \over {49}}} \over {{{\left( {x - r} \right)}^3}}} + \mathop \sum \limits_{r = RootOf\left( {{x^2} - x - 5} \right)} {{ - {r \over {441}} - {4 \over {441}}} \over {{{\left( {x - r} \right)}^2}}} + \mathop \sum \limits_{r = RootOf\left( {{x^2} - x - 5} \right)} {{ - {1 \over {1029}} + {{2r} \over {1029}}} \over {x - r}} \cr} $
2) Integrate the decomposed integrand
Hosein Rahnama
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i think it is not so simple, the printing of the solution of your CAS proves this – Dr. Sonnhard Graubner Dec 02 '15 at 19:53
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@Dr.SonnhardGraubner: Yes, you are right! :) Doing the computation by hand is frustrating in this case! Do we have to the computation by hand? :) Just wanted to give the hint! :) – Hosein Rahnama Dec 02 '15 at 19:55
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@PatrikD: There seem to be no easy way for computing this. The normal procedure is explained in the above answer. :) – Hosein Rahnama Dec 02 '15 at 20:15
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As in my other answer,
$$\int \frac{x+1}{\left(x^2-x-5\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x-5\right)}{\left(x^2-x-5\right)^3}+\int\frac{3}{\left(x^2-x-5\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x-5\right)^2}+3\cdot 2^5\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}\right)$$
Let $2x-1=\sqrt{21}\sin u$. Then $d(2x-1)=\sqrt{21}\cos u\, du$.
$$\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}=\frac{1}{21^2\sqrt{21}}\int \frac{1}{-\cos^5 u}\, du$$
Now integrate by parts. Here are some hints:
$$\int \sec^5 u\, du=\int \sec^3 u\, d\left(\tan u\right)=\sec^3 u\tan u-\int \left(3\sec^2 u\cdot \sec u\tan u\right)\tan u\, du$$
$$\int \sec^3 u\tan^2 u\, du=\int \sec^3 u\left(\sec^2 u-1\right)\, du=\int \sec^5 u\, du-\int \sec^3 u\, du$$
$$\int \sec^3 u\, du=\int \sec u\, d(\tan u)=\sec u\tan u-\int (\sec u\tan u)(\tan u)\, du$$
$$\int \sec u\tan^2 u\, du=\int \sec\left(\sec^2 u-1\right)\, du=\int \sec^3 u\, du-\int \sec u\, du$$
Now, there are many ways to solve $\int \sec u\, du$. See Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ (including the question body). You can always use the Tangent Half-Angle substitution on rational $\sin x,\cos x$ integrals, including $\int \sec u\, du$.
user236182
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