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How should i solve this integral? i know that it is the same question like here Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$ but I've tried solve it for more then 3 hours and i still have no idea ho to solve it. Thank for help. $$\int\frac{x+1}{(x^2+7x-3)^3}dx$$

I tried use $$2x+7 = \frac{\sqrt{61}}{\cos u}$$

$$\int\frac{\frac{\sqrt{61}}{\cos u}-\frac72+1} {(\frac{(\frac{\sqrt{61}}{\cos u})^2-61}{4})^3}dx$$

Now i have $$\int \frac{61\sin u - 5\sqrt{61}\sin u \cos u}{4\cos u \frac{226981\tan^6u}{65} }$$

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  • $\begingroup$ See if this helps: $$\int\frac{x+1}{(x^2+7x-3)^3}dx=\frac 12\int \frac{2x+7}{(x^2+7x-3)^3}dx - \frac 52\int \frac{1}{(x^2+7x-3)^3}dx \\ = \frac 12\int \frac{\frac{d}{dx}(x^2+7x-3)}{(x^2+7x-3)^3}dx - \frac 52\int \frac{1}{\big((x+\frac 72)^2-\frac{61}4\big)^3}dx$$ $\endgroup$ – user137731 Dec 2 '15 at 20:16
  • $\begingroup$ @ Bye_World: Really a nice hint $\endgroup$ – Bhaskara-III Dec 2 '15 at 21:14
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HINT:

$$\int\frac{x+1}{(x^2+7x-3)^3}\space\text{d}x=$$ $$\int\left(\frac{2x+7}{2(x^2+7x-3)^3}-\frac{5}{2(x^2+7x-3)^3}\right)\space\text{d}x=$$ $$\frac{1}{2}\int\frac{2x+7}{(x^2+7x-3)^3}\space\text{d}x-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$


Substitute $u=x^2+7x-3$ and $\text{d}u=(2x+7)\space\text{d}x$:


$$\frac{1}{2}\int\frac{1}{u^3}\space\text{d}u-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$ $$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$ $$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{\left(\left(x+\frac{7}{2}\right)^2-\frac{61}{4}\right)^3}\space\text{d}x=$$


Substitute $s=x+\frac{7}{2}$ and $\text{d}s=\text{d}x$:


$$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{\left(s^2-\frac{61}{4}\right)^3}\space\text{d}s$$




EDIT:

$$\int\frac{1}{(x^2-a)^3}\space\text{d}x=$$


Substitute $x=\sqrt{a}\sec(u)$ and $\text{d}x=\sqrt{a}\tan(u)\sec(u)\space\text{d}u$ so $u=\sec^{-1}\left(\frac{x}{\sqrt{a}}\right)$:


$$\sqrt{a}\int\frac{\cot^4(u)\csc(u)}{a^2}\space\text{d}u=$$ $$\frac{1}{a^{\frac{5}{2}}}\int\cot^4(u)\csc(u)\space\text{d}u=$$ $$\frac{1}{a^{\frac{5}{2}}}\int\csc(u)(\csc^2(u)-1)^2\space\text{d}u=$$ $$\frac{1}{a^{\frac{5}{2}}}\int\left(\csc^5(u)-2\csc^3(u)+\csc(u)\right)\space\text{d}u$$

From now on you can fix it!

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  • $\begingroup$ Thanks a lot, but for me the biggest problem is integrate $\frac{1}{(x^2-a)^3}$ what should i do? $\endgroup$ – DavidM Dec 2 '15 at 20:31
  • $\begingroup$ @DavidM You're welcome! See my edit $\endgroup$ – Jan Dec 2 '15 at 20:49
  • $\begingroup$ really thanks, it is really helpful for me! In edit you show $\frac{1}{(x^2-a)^2}$ what I should do if i have there $()^3$ ? Partial fractions? $\endgroup$ – DavidM Dec 2 '15 at 20:58
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    $\begingroup$ @DavidM I was wrong my $()^2$ has to be $()^3$ the rest of the edit was right! Now my answer is correct and complete $\endgroup$ – Jan Dec 2 '15 at 21:02
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HELP for the integrand $\frac{1}{(x^2-a)^3}$

Use the substitution $x = \sqrt{a}\sec(u)$ and $dx = \sqrt{a}\tan(u) \sec(u) du$ so then $(x^2 - a)^3 = (a\sec^2(u) - a)^3 = a^3\tan^6(u) du$.

Moreover you'll have to take in mind that $u = \text{arcsec}\left(\frac{x}{\sqrt{a}}\right)$

Thus you have

$$\sqrt{a}\int\frac{\cot^4(u)\csc(u)}{a^3} du = \frac{1}{a^{\frac{5}{2}}}\int\cot^4(u)\csc(u)$$

Now use the trigonometric identity

$$\csc^2(u) - \cot^2(u) = 1$$

and arranging the integrand you will obtain

$$ \frac{1}{a^{\frac{5}{2}}}\int \csc^5(u) - 2\csc^3(u) + \csc(u) du $$

Then you can integrate term by term using the recurrence formula for the highest powers:

$$\int\csc^m(u) du = -\frac{\cos(u)\csc^{m-1}(u)}{m-1} + \frac{m-2}{m-1}\int \csc^{m-2}(u) du$$

where in your case $m = 5$ and next $m = 3$. Do the math and you will obtain:

$$ \frac{5\cot(u)\csc(u)}{8a^{5/2}} - \frac{\cot(u)\csc^3(u)}{4a^{5/2}} + \frac{3}{8a^{5/2}}\int\csc(u) du $$

The integral of $\csc(u)$ is $-\log(\csc(u) + \csc(u))$

So you'll get in the end:

$$ \frac{5\cot(u)\csc(u)}{8a^{5/2}} - \frac{\cot(u)\csc^3(u)}{4a^{5/2}} + \frac{3}{8a^{5/2}}(-\log(\csc(u) + \csc(u))) du $$

Now the tedious work: substitute back for $u$ to get the result in $x$: Do the math and get:

$$ -\frac{((\cot(\sec^{-1}(\frac{x}{\sqrt{a}}) \csc(\sec^{-1}(\frac{x}{\sqrt{a}}))^3)}{(4 a^{5/2}))} + \frac{(5 \cot(\sec^{-1}(\frac{x}{\sqrt{a}})) \csc(\sec^{-1}(\frac{x}{\sqrt{a}})))}{(8 a^{5/2})} - \frac{3 \log(\cot(\sec^{-1}(\frac{x}{\sqrt{a}})) + \csc(\sec^{-1}(\frac{x}{\sqrt{a}}))))}{(8 a^{5/2})} $$

Anyway, your integral is a pain.

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  • $\begingroup$ Thanks a lot!!! I am little bit confused. I am not sure if bracelet in final are good $\endgroup$ – DavidM Dec 2 '15 at 21:28
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    $\begingroup$ Sorry, I wanted to write $\csc^2(u) - \cot^2(u) = 1$ going to edit $\endgroup$ – Von Neumann Dec 2 '15 at 22:18

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