Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{2}+1}{(x-2)^{2}(x+3)} d x$

Solution:

$I=\int \frac{x^{2}+2}{(x-2)^{2}(x+3)} d x$

$\frac{\mathrm{x}^{2}+2}{(\mathrm{x}-2)^{2}(\mathrm{x}+3)}=\frac{\mathrm{A}}{\mathrm{x}-2}+\frac{\mathrm{B}}{(\mathrm{x}-2)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+3}$

$x^{2}+1=A(x-2)(x+3)+B(x+3)+C(x-2)^{2}$

Put $x=2$

$4+1=B \times 5$

$5=5 B$

$B=\frac{5}{5}=1$

Put $x=-3$

$10=C \times 25$

$C=\frac{10}{25}=\frac{2}{5}$

Equating coefficients of constants

$1=-6 A+3 B+4 C$

$1=-6 A+3+\frac{8}{5}$

$A=\frac{3}{5}$

Thus,

$I=\frac{3}{5} \int \frac{d x}{x-2}-\int \frac{d x}{(x-2)^{2}}-\frac{2}{5} \int \frac{d x}{x+3}$

$I=\frac{3}{5} \log |x-2|-\frac{1}{(x-2)}+\frac{2}{5} \log |x+3|+C$

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