_0^1 x , , ( 1 + x 2 ) ,dx = - Vidyadip

MCQ

$\int\limits_0^1 {x\,\ln \,\left( {1 + \frac{x}{2}} \right)\,dx} =$

✓
$\frac{3}{4}\left( {1 - 2\ln \,\frac{3}{2}} \right)$
B
$\frac{3}{2} - \frac{7}{2}\ln \,\frac{3}{2}$
C
$\frac{3}{4} + \frac{1}{2}\ln \,\frac{1}{{54}}$
D
$\frac{1}{2}\ln \,\frac{{27}}{2} - \frac{3}{4}$

✓

Answer

Correct option: A.

$\frac{3}{4}\left( {1 - 2\ln \,\frac{3}{2}} \right)$

a
$I =$$\int\limits_0^1 {x\,\,\ln \,\,\left( {\frac{{x + 2}}{2}} \right)\,\,dx\,} $ $=$$\int\limits_0^1 {x\,(\,\ln (x + 2)\, - \,\ln 2)\,\,\,\,dx\,} $

$=\int\limits_0^1 {x\,\,\ln (x + 2)dx\,} \,\, - \,\,\ln 2\,\,\int\limits_0^1 {x\,\,dx\,} \,$

$ \Rightarrow$ $\ln (x + 2)\,.\,\left. {\frac{{{x^2}}}{2}} \right]_0^1\,\, - \,\,\int\limits_0^1 {\frac{{{x^2}}}{{x + 2}}\,\,\,dx\, - \,\,\frac{{\ln 2}}{2}} \,$

$=\frac{1}{2}\,\ln \,3\,\,\, - \,\,\,\int\limits_0^1 {\frac{{{x^2} - 4 + 4}}{{x + 2}}\,\,\,dx\, - \,\,\frac{{\ln 2}}{2}} \,$

$\Rightarrow$ $\frac{1}{2}\,\ln \,\frac{3}{2}\,\,\, - \,\,\,\int\limits_0^1 {\left( {(x - 2)\,\,\, + \,\,\frac{4}{{x + 2}}} \right)\,\,\,dx\,} $

now proceed

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