Question

FInd the value of: ${\int }_0^{\infty }\frac{\ln x}{4x^2+4x+1}dx$

• A. $-\ln 2$
• B. $-\frac{1}{2}\ln 2$
• C. $\frac{1}{2}\ln 2$
• D. None of these

Answer 1

The correct option is B $-\frac{1}{2}\ln 2$

$I={\int }_0^{\infty }\frac{lnx}{4x^2+4x+1}dx$

$={\int }_0^{\infty }\frac{lnx}{(2x+1{)}^2}dx$

$\Rightarrow {\int }_0^{\infty }\underset{\left(I\right)}{lnx}\underset{\left(II\right)}{(2x+1{)}^{-2}}dx$

$\Rightarrow lnx.{\int }_0^{\infty }(2x+1{)}^{-2}dx-{\int }_0^{\infty }\left\{\frac{d}{dx}\right(lnx).\int (2x+1{)}^{-2}dx\}dx$

$\Rightarrow {\left[lnx(\frac{1}{2}\cdot \frac{(2x+1{)}^{-2+1}}{(-2+1)})\right]}_0^{\infty }-{\int }_0^{\infty }\{\left(\frac{1}{x}\right)\cdot \frac{1}{2}\cdot \frac{(2x+1{)}^{-2+1}}{(-2+1)}\}dx$

$\Rightarrow {\left[lnx(-\frac{1}{2}\times \frac{1}{(2x+1)})\right]}_0^{\infty }+\frac{1}{2}\left\{\int \frac{1}{x(2x+1)}\right\}dx$

$\Rightarrow {\left[(-lnx\times \frac{1}{2(2x+1)})\right]}_0^{\infty }+\frac{1}{2}\left[{\int }_0^{\infty }\frac{1}{x}dx-{\int }_0^{\infty }\frac{1}{x+\frac{1}{2}}dx\right]$

$\Rightarrow {[-\frac{lnx}{2(2x+1)}]}_0^{\infty }+\frac{1}{2}{[lnx-ln(x+\frac{1}{2})]}_0^{\infty }$

$=[\frac{-ln\infty }{2(2\times \infty +1)}+\frac{ln0}{2(2\times 0+1)}]+\frac{1}{2}[(ln\infty -ln(\infty +\frac{1}{2}))-(ln0-ln(0+\frac{1}{2}))]$

$=0+\frac{1}{2}[0-(-ln\frac{1}{2})]$

$=\frac{1}{2}ln\frac{1}{2}$

$=-\frac{1}{2}ln2$.