Calculating integrals

[ins-]

Calculate:
$\int \ \frac {(x - 1)\sqrt {x^{4} + 2x^{3} - x^{2} + 2x + 1}}{x^{2}(x + 1)}\,dx$

I know a solution to this integral but I think it is an interesting one and I'll be happy to see different approaches.

P.S. In the past someone said my problems are "beautiful" because there is a symmetry in them but I think this time there is only a beauty.

[arqady]

Подстановка

$$x+\frac{1}{x}=t$$

сильно его упрощает.

[ins-]

Please, write the resulting integral after the substitution you mentioned.

[TOTAL]

$$t=x+\frac{1}{x} -1, \;\;\;\;$$

$$\int \ \frac {\sqrt {t^2 -4}}{t+1}\,dt$$

незваный гость писал(а):

:evil:(but the result is different from TOTAL's).

У меня ошибка, исправил.

[ins-]

Very good idea!
I'm wondering how people invent such substitutions.
Is this a difficult integral?
Do you want me to post some strange integrals?
As I said before in this forum there are very experienced people and I'm interested to see some interesting integrals and to learn the ideas for them.

[ins-]

Try these integrals. I don't need all solutions - only hints. What integral(s) do you like most?
Here you may post some difficult and/or interesting integrals like these you like. Enjoy!

1.
$\int \ x\sqrt {\frac {2sin(x^{2} + 1) - sin(2x^{2} + 1)}{2sin(x^{2} + 1) + sin(2x^{2} + 1)}}\,dx$

2.
$\int \ \frac {1}{(x^{2} + a^{2})\sqrt {x^{2} + b^{2}}}\,dx$ (|b|>|a|)

3.
$\int \ \frac {cosx + xsinx}{{(x + cosx)}^{2}}\,dx$

4.
$\int \ \frac {xlnx}{\sqrt {1 - x^{2}}}\,dx$

5.
$\int \ \frac {1}{x\sqrt {x^{4} + x^{2} + 1}}\,dx$ - EDITED

[незваный гость]

ins- писал(а):

I'm wondering how people invent such substitutions.

From observation, that $x^4 +2 x^3 -x^2 +2 x + 1$ has symmetry of coefficients of complimentary powers of $x$. Thus the substitution $t = x+1/x$ looks natural. Indeed, after simplification we get $\int \frac{\sqrt{(t+3)(t-1)}}{t+2}{\rm d} t$, which can be futher simplified with $u = t+1$ (but the result is different from TOTAL's).

[ins-]

незваный гость - you are right. For this example the substitution t=x+1/x is fine I even think about t=x+1/x+c where c is a constant will be ok, but there are lots of "irrational" integrals and "rational" integrals that can be solved with substitutions of the kind: t=x+1/x that are not so intuitive. It was the reason to ask my question.

[antbez]

The forth one is easy. It may be solved by integrating "by parts"
($dV=\frac{xdx} {\sqrt{(1-x^2)}}$

[ins-]

what about the others?

[GAA]

$$\int \frac{\frac{1-\tan^2x/2}{1+\tan^2x/2} +2x\frac{\tan{x/2}}{1+\tan^2x/2}}{\left( \frac{1-\tan^2x/2}{1+\tan^2x/2}+x\right)^2 } dx =$$

$$ \int \frac{1-\tan^2x/2 + 2x\tan{x/2} }{\left(1+x \frac{1+\tan^2x/2}{1-\tan^2x/2}\right)^2 } \frac{1+\tan^2x/2}{(1-\tan^2x/2)^2}dx=$$

$$\int \frac{ d \left(1+x \frac{1+\tan^2x/2}{1-\tan^2x/2}\right)}{\left(1+x \frac{1+\tan^2x/2}{1-\tan^2x/2}\right)^2 }$$

[ins-]

Very good idea! It probably also may be solved without using universal substitution.

EDITED - the last integral were wrong - I'm sorry. Now the statement is correct. I'm not sure if the statement of the first problem is correct. I'll try to never repeat such a situation.

[nefus]

$\int \ \frac {1}{x\sqrt {x^{4} + x^{2} + 1}}\,dx$
Замена $\frac{1}{x}$=t=>I=-$\int \ \frac {1}{t^2\frac{1}{t}\sqrt {\frac{1}{t^4} + \frac{1}{t^2} + 1}}\,dt=-\int \ \frac {t}{\sqrt {t^4 + t^2 + 1}}\,dt=-\frac{1}{2}\int \ \frac {1}{\sqrt {t^4 + t^2 + 1}}\,d(t^2+\frac{1}{2})=$
$=-\frac{1}{2}\int \ \frac {1}{\sqrt {(t^2 + \frac{1}{2})^2 + \frac{3}{4}}}\,d(t^2+\frac{1}{2})=-\frac{1}{2}ln(t^2+\frac{1}{2}+\sqrt {(t^2 + \frac{1}{2})^2 + \frac{3}{4}})+C=$
$=-\frac{1}{2}ln(t^2+\frac{1}{2}+\sqrt {t^4 + t^2 + 1})+C$
Далее обратная замена:
$I=-\frac{1}{2}ln(\frac{1}{x^2}+\frac{1}{2}+\sqrt {\frac{1}{x^4} + \frac{1}{x^2}+1})+C$

[ins-]

2 - may be started with the same substitution. I think it is the most interesting if it is not a "tabular" integral.

[GAA]

substitution $u=\frac{x}{\sqrt{x^2+b^2}}$