Show that

The given differential equation is:

$\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x$

$\Rightarrow \frac{d y}{d x}=\frac{2 x^{2}+x}{\left(x^{3}+x^{2}+x+1\right)}$

$\Rightarrow d y=\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x$

Integrating both sides, we get:

$\int d y=\int \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x$            ...(1)

Let $\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1}$.              ...(2)

$\Rightarrow \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A x^{2}+A+(B x+C)(x+1)}{(x+1)\left(x^{2}+1\right)}$

$\Rightarrow 2 x^{2}+x=A x^{2}+A+B x^{2}+B x+C x+C$

$\Rightarrow 2 x^{2}+x=(A+B) x^{2}+(B+C) x+(A+C)$

Comparing the coefficients of $x^{2}$ and $x$, we get:

A + B = 2

B + C = 1

A + C = 0

Solving these equations, we get:

$A=\frac{1}{2}, B=\frac{3}{2}$ and $C=\frac{-1}{2}$

Substituting the values of A, B, and C in equation (2), we get:

$\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{(3 x-1)}{\left(x^{2}+1\right)}$

Therefore, equation (1) becomes:

$\int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^{2}+1} d x$

$\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x$

$\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \cdot \int \frac{2 x}{x^{2}+1} d x-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$

$\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$

$\Rightarrow y=\frac{1}{4}\left[2 \log (x+1)+3 \log \left(x^{2}+1\right)\right]-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$

$\Rightarrow y=\frac{1}{4}\left[(x+1)^{2}\left(x^{2}+1\right)^{3}\right]-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$               ...(3)

Now, $y=1$ when $x=0$

$\Rightarrow \mathrm{I}=\frac{1}{4} \log (\mathrm{l})-\frac{1}{2} \tan ^{-1} 0+\mathrm{C}$

$\Rightarrow \mathrm{I}=\frac{1}{4} \times 0-\frac{1}{2} \times 0+\mathrm{C}$

$\Rightarrow \mathrm{C}=1$

Substituting C = 1 in equation (3), we get:

$y=\frac{1}{4}\left[\log (x+1)^{2}\left(x^{2}+1\right)^{3}\right]-\frac{1}{2} \tan ^{-1} x+1$