Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$

Question

Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$

$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$

$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \int\frac{3}{(x^2+x+1)^2}dx$$

Now Using the formula $$\bullet x^7-1 = (x-1)\cdot \left[x^6+x^5+x^4+x^3+x^2+x+1\right]$$

So $$\bullet \; (x^7-1) = (x-1)\left[(x^4+x)\cdot (x^2+x+1)+1\right]$$

So we get $$\displaystyle I = \int\frac{(x-1)\cdot (x^4+x)\cdot (x^2+x+1)}{(x^2+x+1)^2}dx+\frac{1}{2}\int\frac{2x-2}{(x^2+x+1)^2}+3\int\frac{1}{(x^2+x+1)^2}dx$$

$$\displaystyle I = \underbrace{\int\frac{(x-1)(x^4+x)}{x^2+x+1}dx}_{J}+\underbrace{\frac{1}{2}\int\frac{(2x+1)}{(x^2+x+1)^2}dx}_{K}+\underbrace{\frac{3}{2}\int\frac{1}{(x^2+x+1)^2}dx}_{L}$$

Now $$\displaystyle J = \int\frac{(x-1)(x^4-x)+2x(x-1)}{(x^2+x+1)}dx = \int(x^3-2x^2+x)dx+\int\frac{2x^2-2x}{x^2+x+1}dx$$

Now we can solve the integral Using the formulae.

My question is can we solve it any other way, If yes then plz explain here

Thanks

Answer 1

HINT: prove that $$\frac{x^7+2}{(x^2+x+1)^2}={x}^{3}-2\,{x}^{2}+x+2+{\frac {-4\,x-2}{{x}^{2}+x+1}}+{\frac {x+2}{ \left( {x}^{2}+x+1 \right) ^{2}}} $$

Answer 2

HINT:

Notice, we have $$\int\frac{x^7+2}{(x^2+x+1)^2}dx$$ $$=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx$$ Let, $$x+\frac{1}{2}=t\implies dx=dt$$

$$=\int\frac{\left(t-\frac{1}{2}\right)^7+2}{\left(t^2+\frac{3}{4}\right)^2}dt$$

Use reduction to solve further

Answer 3

Here is a sketch - not a full solution:

Say we know $\int\cos^2x\tan^jx dx$ for $j=0,1,...,7$. Then using $x+\frac12=\frac{\sqrt3}{2}\tan u$ we have \begin{align} \int\frac{x^7+2}{(x^2+x+1)^2}dx&=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx\\ &=\frac{\sqrt3}{2}\int\frac{\Big(\frac{\sqrt3}{2}\tan u-\frac12\Big)^7+2}{\frac{9}{16}(1+\tan^2u)}du\\ &=\frac{\sqrt3}{2}\frac{32}{9}\int\cos^2udu+\frac{\sqrt3}{2}\sum_{j=0}^7\Big(-\frac12\Big)^j\Big(\frac{\sqrt3}{2}\Big)^{7-j}\frac{16}{9}\int\cos^2u\tan^{7-j}udu \end{align}