Question

If ${\int }_0^{40}\frac{dx}{2x+1}=\log a,$ then $a$ is:

• A. $3$
• B. $9$
• C. $81$
• D. $40$

Answer 1

Answer: (B)

${\int }_0^{40}\frac{dx}{2x+1}$

$={\left[\frac{\log (2x+1)}{2}\right]}_0^{40}$ [ $\because \int \frac{1}{ax+b}=\frac{\log (ax+b)}{a}+C$]

$=\frac{1}{2}\left[\log \right(2\left(40\right)+1)-\log (2\left(0\right)+1\left)\right]$

$==\frac{1}{2}[\log 81-\log 1]$

$=\frac{1}{2}\log 81$

$=\log {81}^{\frac{1}{2}}$

$=\log \sqrt{81}$

$=\log 9$

$\therefore {\int }_0^{40}\frac{dx}{2x+1}=\log 9$

It is given that, ${\int }_0^{40}\frac{dx}{2x+1}=\log a$

Thus the value of $a$ is $9$.

Hence, Option A is correct.