$({x^3} + {x^2} + x + 1)\cfrac{{dy}}{{dx}} = 2{x^2} + x;y = 1$ when $x = 0$

.: We have, $({x^3} + {x^2} + x + 1)\cfrac{{dy}}{{dx}} = 2{x^2} + x$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{2{x^2} + x}}{{{x^3} + {x^2} + x + 1}}$

$\Rightarrow dy = \cfrac{{2{x^2} + x}}{{{x^2}(x + 1) + 1(x + 1)}}dx$

… (1)
Integrating (1) both sides,

we get
$\int d y = \int {\cfrac{{2{x^2} + x}}{{({x^2} + 1)(x + 1)}}dx}$ $\Rightarrow y = \int {\cfrac{{2{x^2} + x}}{{({x^2} + 1)(x + 1)}}dx}$

Now, let $\cfrac{{2{x^2} + x}}{{({x^2} + 1)(x + 1)}} = \cfrac{A}{{x + 1}} + \cfrac{{Bx + C}}{{{x^2} + 1}}$

$\Rightarrow 2{x^2} + x = A({x^2} + 1) + (Bx + C)(x + 1)$

Comparing coefficients of $x$,

we get
$1 = B + C$ … (2)
Comparing constant terms,

we get
$0 = A + C \Rightarrow C = - A$ … (3)
Comparing coefficients of ${x^2}$,

we get
$2 = A + B$ … (4)
Now solving (2) \& (3),

we get
$- A + B = 1$ … (5)
Solving (4) \& (5), we get $2B = 3 \Rightarrow B = 3/2$

Substituting the value of $B$ in (2) \& (5),

we get
$A = 1/2$ and $C = - 1/2$

$\therefore$ $y = \int {\left[ {\cfrac{{\cfrac{1}{2}}}{{x + 1}} + \cfrac{{\cfrac{3}{2}x - \cfrac{1}{2}}}{{{x^2} + 1}}} \right]dx}$

$\Rightarrow y = \cfrac{1}{2}\log (x + 1) + \cfrac{3}{4}\int {\cfrac{{2x}}{{{x^2} + 1}}} dx - \cfrac{1}{2}\int {\cfrac{{dx}}{{{x^2} + 1}}}$

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

When $x = 0,y = 1$

$\Rightarrow 1 = \cfrac{1}{2}\log 1 + \cfrac{3}{4}\log 1 - \cfrac{1}{2}{\tan ^{ - 1}}(0) + C \Rightarrow C = 1$

Hence, particular solution is

$y = \cfrac{1}{2}\log (x + 1) + \cfrac{3}{4}\log ({x^2} + 1) - \cfrac{1}{2}{\tan ^{ - 1}}x + 1$

$\Rightarrow y = \cfrac{1}{4}\log {(x + 1)^2} + \cfrac{1}{4}\log {({x^2} + 1)^3} - \cfrac{1}{2}{\tan ^{ - 1}}x + 1$

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