<s>[INST] Evaluation of $ \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$ 
Evaluation of $\displaystyle \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$

$\bf{My\; Try::}$ We can write $$1+2x+3x^2+4x^3+3x^4+2x^5+x^6 $$
$$= (1+x)+(x+x^2)+2(x^2+x^3)+2(x^3+x^4)+(x^4+x^5)+(x^5+x^6)$$
$$ = (1+x)\left[1+x+2x^2+2x^3+x^4+x^5\right]$$
So Integral $$\displaystyle I = \int\frac{[(1-x+x^2)\cdot (1+x+x^2)+x^2]}{1+x+2x^2+2x^3+x^4+x^5}dx$$
So $$\displaystyle I = \int\frac{(1+x^2)^2}{1+x+2x^2+2x^3+x^4+x^5}dx$$
Now How can I solve it, Help me
Thanks.
 [/INST] Well, you're really almost done. All you have to do is realize that the denominator factors more: $$1+x+2x^2+2x^3+x^4+x^5=(1+x)(1+x^2)^2$$
This of course gives you
$$I=\int\frac{1}{1+x}\,dx=\ln|1+x|+C$$
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