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  4. Evaluate: ∫(3x+1/(x-2)2(x+2))dx

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Q. Evaluate : $\int\frac{3x+1}{\left(x-2\right)^{2}\left(x+2\right)}dx$

Integrals

Solution:

Let $\frac{3x+1}{\left(x-2\right)^{2}\left( x+2\right)} = \frac{A}{x-2} + \frac{B}{\left(x-2\right)^{2} }+\frac{C}{ x+2} ............\left(i\right)$

$\Rightarrow 3x+1 =A\left(x-2\right)\left(x+2\right) +B\left(x+2\right)+ C\left(x-2\right)^{2} .....\left(ii\right)$

Put $x = 2$ in (ii), we get $7 = 4B \Rightarrow B =\frac{7}{4}$

Put $x = -2$ in (ii), we get $-5 = 16 C \Rightarrow C = -\frac{5}{16}$

Comparing coefficients of $x^{2}$ on both sides of $\left(ii\right)$, we get $A + C = 0\Rightarrow A = -C \Rightarrow A =\frac{5}{16}$

$ \therefore I= \int \frac{3x+1}{\left(x-2\right)^{2}\left(x+2\right)}dx$

$ =\frac{5}{16}\int\frac{1}{x-2}dx +\frac{7}{4}\int\frac{1}{\left(x-2\right)^{2}}dx -\frac{5}{16}\int\frac{1}{x+2}dx $

$ \Rightarrow I= \frac{5}{16}log\left|x-2\right| -\frac{7}{4\left(x-2\right)}-\frac{5}{16}log\left|x+2\right| +C$