Let $I =\int \frac{dx}{x^3+3x^2+2x}$

Factorizing the denominator of the integral $I$, we get

$\Rightarrow I=\int \frac{dx}{x\left(x+1\right)\left(x+2\right)}$ 

$\Rightarrow I=\int \left(\frac{1}{x+1}-\frac{1}{x+2}\right)\frac{1}{x}dx$

$\Rightarrow I=\int \frac{1}{x\left(x+1\right)}dx-\int \frac{1}{x\left(x+2\right)}dx$

$\Rightarrow I=\int \left(\frac{1}{x}-\frac{1}{x+1}\right)dx-\frac{1}{2}\int \left(\frac{1}{x}-\frac{1}{x+2}\right)dx$

Using $\int \frac{1}{x}dx=\log \left|x\right|$

$\Rightarrow I=\log \left|x\right|-\log \left|x+1\right|-\frac{1}{2}\log \left|x\right|+\frac{1}{2}\log \left|x+2\right|+c$

$\Rightarrow I=\frac{1}{2}\left[\log \left|x\right|+\log \left|x+2\right|-2\log \left|x+1\right|\right]+c$

Using $\log \left(m^n\right)=n\log m$

$\Rightarrow I=\frac{1}{2}\left[\log \left|x\right|+\log \left|x+2\right|-\log {\left(x+1\right)}^2\right]+c$

Using $\log m+\log n=\log \left(mn\right) \& \log m-\log n=\log \left(\frac{m}{n}\right)$

The required solution is $I=\frac{1}{2}\log \left[\frac{\left|x^2+2x\right|}{{\left(x+1\right)}^2}\right]+c.$

Let the centre of one such circle be $P(h,k)$ since it touches $y-axis$ in the first quadrant
$\Rightarrow$ its radius $=h$
Now, since it touches $x^2+y^2=1$ externally
$\Rightarrow h+1=\sqrt{h^2+k^2}$
$\Rightarrow h^2+2h+1=h^2{k}^2$
$\Rightarrow k^2=2h+1$
Hence the required locus is
$y^2=2x+1$
$\Rightarrow \mathrm{y}=\mathrm{ }\sqrt{1+2\mathrm{x}},\mathrm{ }\mathrm{ }\mathrm{x}\ge 0$