Let $y=y\left(x\right)$ be the solution curve of the differential equation $\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{\left(x+3\right)}{x+1},x>-1$, which passes through the point $\left(0,1\right)$. Then $y\left(1\right)$ is equal to

Given,

$\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1}$

Now comparing with $\frac{dy}{dx}+y\times p\left(x\right)=q\left(x\right)$
We get $p\left(x\right)=\frac{2x^2+11x+13}{x^3+6x^2+11x+6}$

So $IF=e^{\int \frac{2x^2+11x+13}{x^3+6x^2+11x+6}dx}$

Now finding the integration we get,

$\int p\left(x\right)dx=\int \frac{\left(2x^2+11x+13\right)dx}{\left(x+1\right){\displaystyle \left(x+2\right)}{\displaystyle \left(x+3\right)}}$

Using partial fraction we get,

$\frac{2x^2+11x+13}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$

On solving we get $A=\frac{4}{2}=2$, $B=1$ and $C=-1$

So, $\int p\left(x\right)dx=A\ln \left(x+1\right)+B\ln \left(x+2\right)+c\ln \left(x+3\right)$

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

So, $IF=e^{\int p\left(x\right)dx}=\frac{{\left(x+1\right)}^2\left(x+2\right)}{\left(x+3\right)}$

Now solution of differential equation is given by

 $y\times IF=\int Q\times IFdx$

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

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

Now given curve passes through $\left(0,1\right)$,

So, $\Rightarrow 1\times \left(\frac{{\displaystyle {\left(0+1\right)}^2\left(0+2\right)}}{0+3}\right)=\frac{0^3}{3}+\frac{3\times 0^2}{2}+2\times 0+c\Rightarrow c=\frac{2}{3}$

So, curve becomes $y\times \left(\frac{{\displaystyle {\left(x+1\right)}^2\left(x+2\right)}}{x+3}\right)=\frac{x^3}{3}+\frac{3x^2}{2}+2x+\frac{2}{3}$

Now put $x=1$ we get, $y\times \left(\frac{{\displaystyle {\left(1+1\right)}^2\left(1+2\right)}}{1+3}\right)=\frac{1^3}{3}+\frac{3\times 1^2}{2}+2\times 1+\frac{2}{3}$

$\Rightarrow y\times 3=1+\frac{3}{2}+2\Rightarrow y\times 3=\frac{9}{2}$

 $\Rightarrow y\left(1\right)=\frac{3}{2}$