If $\int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}dx=\frac{\mathrm{A}}{x-1}+B \log \left(2\mathrm{x}-1\right)+C \log \left(\mathrm{x}-1\right)+K$ then $A+B+C=$

Given,

$\int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}dx=\frac{\mathrm{A}}{x-1}+B \log \left(2\mathrm{x}-1\right)+C \log \left(\mathrm{x}-1\right)+K$

Now solving $\frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}$ by partial fraction we get,

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

$\Rightarrow x+3=\alpha \left(x-1\right)\left(2x-1\right)+\beta \left(2x-1\right)+\gamma {\left(x-1\right)}^2$

Now comparing coefficient of $x^2, x \& \text{constant term}$ we get,

$\alpha =-7, \beta =4 \& \gamma =14$

Now putting the value and integrating the given function we get,

$\int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}=\int \frac{-7}{x-1}+\int \frac{4}{{\left(x-1\right)}^2}+\int \frac{14}{\left(2x-1\right)}$

$\Rightarrow \int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}=-7\log \left(x-1\right)+\frac{-4}{\left(x-1\right)}+7\log \left(2x-1\right)+K$

Now comparing with $\int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}dx=\frac{\mathrm{A}}{x-1}+B \log \left(2\mathrm{x}-1\right)+C \log \left(\mathrm{x}-1\right)+K$

We get,

$A=-7, B-4 \& C=7$

Then $A+B+C=-4$