Question

# Q=\int_4^9\frac{3x+12}{x^2+6x+9}dx Determine the numerical values of the coefficients, A and B, where A\leq B. \frac{A}{denomianator}+\frac{B}{denomiantor} Need to find A and B.

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$$\displaystyle{Q}={\int_{{4}}^{{9}}}{\frac{{{3}{x}+{12}}}{{{x}^{{2}}+{6}{x}+{9}}}}{\left.{d}{x}\right.}$$
Determine the numerical values of the coefficients, A and B, where $$\displaystyle{A}\leq{B}$$.
$$\displaystyle{\frac{{{A}}}{{{d}{e}{n}{o}{m}{i}{a}{n}{a}\to{r}}}}+{\frac{{{B}}}{{{d}{e}{n}{o}{m}{i}{a}{n}\to{r}}}}$$
Need to find A and B.

2021-04-08

We take:
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{3}{x}+{12}}}{{{\left({x}+{3}\right)}^{{2}}}}}$$
we write
$$\displaystyle{\frac{{{A}}}{{{x}+{3}}}}+{\frac{{{B}}}{{{\left({x}+{3}\right)}^{{2}}}}}={\frac{{{3}{x}+{12}}}{{{\left({x}+{3}\right)}^{{2}}}}}$$
$$\displaystyle{A}{\left({x}+{3}\right)}+{B}={3}{x}+{12}$$ we put x=-3 and get B=3
we put x=0 we get
3A+B=12 put b=3 we get A=3
Then we have
$$\displaystyle{\frac{{{3}{x}+{12}}}{{{x}^{{2}}+{6}{x}+{9}}}}={\frac{{{3}}}{{{\left({x}+{3}\right)}}}}+{\frac{{{3}}}{{{\left({x}+{3}\right)}^{{2}}}}}$$
$$Q=\int_4^9\frac{3x+12}{x^2+6x+9}dx$$
$$\displaystyle={3}{\int_{{4}}^{{9}}}{\left[{\frac{{{1}}}{{{x}+{3}}}}+{\left({x}+{3}\right)}^{{-{2}}}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle={3}{{\left[{\ln{{\left({x}+{3}\right)}}}-{\frac{{{1}}}{{{x}+{3}}}}\right]}_{{4}}^{{9}}}$$
$$\displaystyle={3}{\left[{\ln{{12}}}-{\frac{{{1}}}{{{12}}}}\right]}-{\ln{{7}}}+{\frac{{{1}}}{{{7}}}}{]}$$
$$\displaystyle={3}{\ln{{\left({\frac{{{12}}}{{{7}}}}\right)}}}+{\frac{{{5}}}{{{28}}}}$$