Question

If $\int \frac{x+2}{2x^2+6x+5}dx=P\int \frac{4x+6}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{dx}{2x^2+6x+5}$, then the value of $P$ is:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $2$
• E. $1$

Answer 1

The correct option is C

$\frac{1}{4}$

Explanation for the correct option.

In $\int \frac{x+2}{2x^2+6x+5}dx$,

$x+2$ can be written as $A\frac{d}{dx}\left(2x^2+6x+5\right)+B$

$\begin{array}{rcl}& \Rightarrow & x+2=A\left(4x+6\right)+B\\ & =& 4Ax+6A+B\end{array}$

From here, we get

$\begin{array}{rcl}4A& =& 1\\ & \Rightarrow & A=\frac{1}{4}\end{array}$ and

$$\begin{gathered} 2=6A+B \\ \Rightarrow B=\frac{1}{2} \end{gathered}$$

So,

$\begin{array}{rcl}\int \frac{x+2}{2x^2+6x+5}dx& =& \int \frac{{\displaystyle \frac{1}{4}}\left(4x+6\right)+{\displaystyle \frac{1}{2}}}{2x^2+6x+5}dx\\ & =& \frac{1}{4}\int \frac{{\displaystyle 4x+6}}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{{\displaystyle dx}}{2x^2+6x+5}\end{array}$

By substituitng this value in the given equation, we get

$\begin{array}{rcl}& & \frac{1}{4}\end{array}\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & \frac{{\displaystyle 4x+6}}{2x^2+6x+5}\end{array}\begin{array}{l}d\end{array}\begin{array}{l}x\end{array}\begin{array}{l}+\end{array}\begin{array}{l}\frac{1}{2}\end{array}\begin{array}{l}\int \end{array}\frac{{\displaystyle dx}}{2x^2+6x+5}\begin{array}{rcl}& =& \end{array}P\int \frac{4x+6}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{dx}{2x^2+6x+5}$

Therefore, $P=\frac{1}{4}$

Hence, option C is correct.