Question

$\int_{}^{}\frac{3x + 1}{(x - 2)^{2}(x + 2)} =$

• A. $\frac{\mathbf{5}}{\mathbf{16}}\mathbf{\log}\left| \frac{\mathbf{x + 2}}{\mathbf{x - 2}} \right|\mathbf{+}\frac{\mathbf{7}}{\mathbf{4(x - 2)}}\mathbf{+ c}$
• B. $\frac{\mathbf{5}}{\mathbf{16}}\mathbf{\log}\left| \frac{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{x + 2}} \right|\mathbf{+}\frac{\mathbf{7}}{\mathbf{4(x}\mathbf{-}\mathbf{2)}}\mathbf{+ c}$
• C. $\frac{\mathbf{16}}{\mathbf{5}}\mathbf{\log}\left| \frac{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{x + 2}} \right|\mathbf{-}\frac{\mathbf{7}}{\mathbf{4(x}\mathbf{-}\mathbf{2)}}\mathbf{+ c}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\mathbf{5}}{\mathbf{16}}\mathbf{\log}\left| \frac{\mathbf{x}\mathbf{-}\mathbf{2}}{\mathbf{x + 2}} \right|\mathbf{+}\frac{\mathbf{7}}{\mathbf{4(x}\mathbf{-}\mathbf{2)}}\mathbf{+ c}$

Answer 2

We have, $\frac{3x + 1}{(x - 2)^{2}(x + 2)} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^{2}} + \frac{C}{(x + 2)}3x + 1 = A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^{2}$ ......(i)

Putting $x = 2\text{and-2}$successively in equation (i), we get

$B = \frac{7}{4},C = \frac{- 5}{16}$

Now, we put $x = 0$ and get $A = \frac{5}{16}$

$\int_{}^{}\frac{3x + 1}{(x - 2)^{2}(x + 2)}dx$ $= \frac{5}{16}\int_{}^{}{\frac{dx}{x - 2} + \frac{7}{4}\int_{}^{}{\frac{dx}{(x - 2)^{2}}\frac{- 5}{16}\int_{}^{}\frac{dx}{x + 2}}} = \frac{5}{16}\log(x - 2) - \frac{7}{4}\frac{1}{(x - 2)} - \frac{5}{16}\log(x + 2) + c$ $= \frac{5}{16}\log\left| \frac{x - 2}{x + 2} \right| ⥂ - \frac{7}{4(x - 2)} + c$