Question

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then

विकल्प

• \[ a = - \frac{1}{10}, b = - \frac{2}{5}\]

• \[a = \frac{1}{10}, b = - \frac{2}{5}\]

• \[ a = - \frac{1}{10}, b = \frac{2}{5}\]

• \[ a = \frac{1}{10}, b = \frac{2}{5}\]

Answer 1

\[ a = - \frac{1}{10}, b = \frac{2}{5}\]

 

\[\text{Let }I = \int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx\]
We express,
\[\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow 1 = A\left( x^2 + 1 \right) + \left( Bx + C \right)\left( x + 2 \right)\]
On comparing the coefficients of `x^2, x` and constants, we get
\[0 = A + B\text{ and }0 = 2B + C\text{ and }1 = A + 2C\]
\[\text{or }A = \frac{1}{5}\text{ and }B = - \frac{1}{5}\text{ and }C = \frac{2}{5}\]
\[ \therefore I = \int\left( \frac{\frac{1}{5}}{x + 2} + \frac{- \frac{1}{5}x + \frac{2}{5}}{x^2 + 1} \right)dx\]
\[ = \frac{1}{5}\int\frac{1}{x + 2}dx - \frac{1}{5}\int\frac{x}{x^2 + 1}dx + \frac{2}{5}\int\frac{1}{x^2 + 1}dx\]
\[ = \frac{1}{5}\log\left| x + 2 \right| - \frac{1}{10}\log\left| x^2 + 1 \right| + \frac{2}{5} \tan^{- 1} x + C\]
\[\text{Since, }\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C\]
\[\text{Therefore, }a = - \frac{1}{10}\text{ and }b = \frac{2}{5}\]