Помогите, пожалуйста, решить интеграл

Question 1

Question 2

Решите задачу:

$\int \dfrac{x^2+x}{x^6+1}\, dx=\int \dfrac{x^2\, dx}{(x^3)^2+1}+\int \dfrac{x\, dx}{(x^2)^3+1}=\underbrace {\dfrac{1}{3}\int \dfrac{d(x^3)}{(x^3)^2+1}}_{t=x^3}+\dfrac{1}{2}\int \dfrac{d(x^2)}{(x^2)^3+1}=\\\\\\=\dfrac{1}{3}\cdot arctg(x^3)+\dfrac{1}{2}\int \dfrac{d(x^2)}{(x^2)^3+1}=\dfrac{1}{3}\cdot arctg(x^3)+\dfrac{1}{2}\, Q\; ;$

$\star \; \; Q=\int \dfrac{d(x^2)}{(x^2)^3+1}=\Big[\; t=x^2\; ,\; dt=d(x^2)\; \Big]=\int \dfrac{dt}{t^3+1}=\\\\=\int \dfrac{dt}{(t+1)(t^2-t+1)}=\int \dfrac{A\, dt}{t+1}+\int \dfrac{(Bt+C)\, dt}{t^2-t+1}=\dfrac{1}{3}\int \dfrac{dt}{t+1}-\dfrac{1}{3}\int \dfrac{(t-2)\, dt}{t^2-t+1}=\\\\\\=\dfrac{1}{3}\, ln|t+1|-\dfrac{1}{3}\int \dfrac{(t-2)\, dt}{(t-\frac{1}{2})^2+\frac{3}{4}}\, dt=\dfrac{1}{3}\, ln|t+1|-\dfrac{1}{3}\int \dfrac{(t-\frac{1}{2})-\frac{3}{2}}{(t-\frac{1}{2})^2+\frac{3}{4}}\, dt=$

$=\dfrac{1}{3}\, ln|t+1|-\dfrac{1}{6}\cdot ln|(t-\frac{1}{2})^2+\frac{3}{4}|+\dfrac{1}{2}\cdot \dfrac{2}{\sqrt3}\cdot arctg\dfrac{2(t-\frac{1}{2})}{\sqrt3}+C=\\\\\\=\dfrac{1}{3}\, ln|x^2+1|-\dfrac{1}{6}\, ln|t^2-t+1|+\dfrac{1}{\sqrt3}\cdot arctg\dfrac{2t-1}{\sqrt3}+C=\\\\\\=\dfrac{1}{3}\, ln|x^2+1|-\dfrac{1}{6}\, ln|x^4-x^2+1|+\dfrac{1}{\sqrt3}\cdot arctg\dfrac{2x^2-1}{\sqrt3}+C$

$\int \dfrac{x^2+x}{x^6+1}\, dx=\\\\\\=\dfrac{1}{3}arctg(x^3)+\dfrac{1}{6}\, ln|x^2+1|-\dfrac{1}{12}\, ln|x^4-x^2+1|+\dfrac{1}{2\sqrt3}\cdot arctg\dfrac{2x^2-1}{\sqrt3}+C_1$