This question was by Hadjer2015 Difficult Integration

I looked for a way to get the sin and cos terms the "same" in some sense, and I could see that if we had integer multiples of some common angle we could expand the result to get both in terms of the one angle.  

There might well be a more straightforward way of doing the integral!

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$$\\\displaystyle\int \frac{x^2}{(x^2+3)(x+1)}\;dx\\\\
Let \\
\frac{x^2}{(x^2+3)(x+1)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+1}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(Ax+B)(x+1)}{(x^2+3)(x+1)}+\frac{C(x^2+3)}{(x^2+3)(x+1)}\\\\
so\\
(Ax+B)(x+1)+C(x^2+3)=x^2\\\\
Ax^2+Ax+Bx+B+Cx^2+3C=x^2\\\\
(A+C)x^2+(A+B)x+(B+3C)=x^2\\\\$$

$$\\so\\
A+C=1 \;(1)\;\rightarrow \\
A+B=0\;(2)\; \rightarrow \\
B+3C=0\;\; \rightarrow\;\;B=-3C\;\; (3) \;\;$sub into (2)$\\
A-3C=0 (4)\\
(1)-(4)\\
4C=1\\
C=\frac{1}{4}=0.25 \\
A=1-0.25=0.75\\
B=-3*0.25=-0.75\\\\$$

$$\\so\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)(x+1)}{(x^2+3)(x+1)}+\frac{0.25(x^2+3)}{(x^2+3)(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{0.75x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{0.375*2x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\\\
\displaystyle\int\;\frac{x^2}{(x^2+3)(x+1)}\;dx\\\\\\
=\displaystyle\int\;\frac{0.375*2x}{(x^2+3)}\;dx-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx+\displaystyle\int\;\frac{0.25}{(x+1)}\;dx\\\\\\
=0.375\;ln(x^2+3)+0.35\;ln(x+1)-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx\\\\\\
$ Now I don't know what to do. BUMMER!$$$

I guess I wasn't meant to do it that way afterall.      LOL

Melody....for that last integral.....let u = x , a = √3,   du = dx,  and we have the form.....

-0.75 ∫ 1 / [ u^2 + a^2] du  =

-0.75 [ tan^-1 (u/a)] / a  + C  =

-.0.75 [tan ^-1 (x / √3)] / √3 + C

Thanks Chris, I should have recognised that. :)

Well...you got most of the difficult part done with the partial fraction stuff....!!!!

That's worth at least 3 points.........

P.S.  - I don't want to appear as though I'm a genius.....I went to the integral tables to find the proper form for that last part.....LOL!!!

For the first integral:

This should be carefully checked as I tend to be a little cavalier with constants sometimes!

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Thanks Alan. :)

Nice "double-substitution,"  Alan.

P.S -  How did it occur to you to let cos(x/3)  = cos(2Θ)   and sin(x/ 2) = sin(3Θ)  ???