Question

Suppose $A=\int \frac{dx}{x^2+6x+25}$ and $B=\int \frac{dx}{x^2-6x-27}$. If $12(A+B)=\lambda .{\tan }^{-1}\left(\frac{x+3}{4}\right)+\mu .\ln \left|\frac{x-9}{x+3}\right|+C$, then the value of $(\lambda +\mu )$ is $\dots$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Given $12(A+B)=\lambda .{\tan }^{-1}\left(\frac{x+3}{4}\right)+\mu .\ln \left|\frac{x-9}{x+3}\right|+C$ .....(1)
Given $A=\int \frac{dx}{x^2+6x+25}$
$A=\int \frac{dx}{(x+3{)}^2+4^2}$
Put $x+3=t$
$dx=dt$
$\Rightarrow A=\int \frac{dt}{t^2+4^2}$
$A=\frac{1}{4}{\tan }^{-1}\frac{t}{4}+c_1$
$\Rightarrow A=\frac{1}{4}{\tan }^{-1}\frac{x+3}{4}+c_1$
Given $B=\int \frac{dx}{x^2-6x-27}$
$B=\int \frac{dx}{(x-3{)}^2-6^2}$
Put $x-3=u$
$dx=du$
$B=\int \frac{du}{u^2-6^2}$
$\Rightarrow B=\frac{1}{12}\ln |\frac{t-6}{t+6}|+c_2$
$\Rightarrow B=\frac{1}{12}\ln |\frac{x-9}{x+3}|+c_2$
Now, $12(A+B)=12[\frac{1}{4}{\tan }^{-1}\frac{x+3}{4}+\frac{1}{2.6}\ln \left|\frac{x-9}{x+3}\right|]=3{\tan }^{-1}\left(\frac{x+3}{4}\right)+\ln \left|\frac{x-9}{x+3}\right|$
Put this value in (1), we get
$\lambda =3$, $\mu =1⟹\lambda +\mu =4$