Question

Solve $\underset{0}{\overset{1}{\int }}x\ln \left(1+\frac{x}{2}\right)dx$

• A. $I=\frac{3}{4}(1-\ln \left(\frac{3}{2}\right))$
• B. $I=-\frac{3}{4}(1-\ln \left(\frac{3}{2}\right))$
• C. $I=\frac{3}{4}(1+\ln \left(\frac{3}{2}\right))$
• D. None of these

Answer 1

The correct option is A $I=\frac{3}{4}(1-\ln \left(\frac{3}{2}\right))$

We have,

${\int }_0^{1}x\ln (1+\frac{x}{2})dx$

$={\int }_0^1\ln (1+\frac{x}{2})\times xdx$

We know that,

$\int u.v=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

Then,

${\int }_0^1\ln (1+\frac{x}{2})\times xdx=ln(1+\frac{x}{2}){\int }_0^{1}xdx-{\int }_0^1\left(\frac{d}{dx}ln(1+\frac{x}{2}){\int }_0^{1}xdx\right)dx$

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

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

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

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{\int }_0^1\left(\frac{x^2}{(2+x)}\right)dx$

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{\int }_0^1\left(\frac{x^2-4+4}{(2+x)}\right)dx$

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{\int }_0^1(\frac{x^2-4}{(2+x)}+\frac{4}{(x+2)})dx$

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{\int }_0^1((x-2)+\frac{4}{(x+2)})dx$

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{\int }_0^{1}xdx+\frac{1}{2}{\int }_0^{1}2dx-\frac{1}{2}{\int }_0^1\frac{4}{(x+2)}dx$

$=ln(1+\frac{x}{2}){{\left[\frac{x^2}{2}\right]}_0}^1-\frac{1}{2}{{\left[\frac{x^2}{2}\right]}_0}^1+\frac{2}{2}{{\left[x\right]}_0}^1-\frac{4}{2}{{\left[\ln (x+2)\right]}_0}^1$

$={{\left[ln(1+\frac{x}{2})\right]}_0}^{1}1[\frac{1^2}{2}-\frac{0^2}{2}]-\frac{1}{4}[1^2-0^2]+[1-0]-2[\ln \left(3\right)-\ln \left(2\right)]$

$=[\ln \left(\frac{3}{2}\right)-\ln \left(1\right)]\times 1[\frac{1^2}{2}-\frac{0^2}{2}]-\frac{1}{4}[1^2-0^2]+[1-0]-2[\ln \left(3\right)-\ln \left(2\right)]$

$=\frac{1}{2}\left[\ln \frac{\left(\frac{3}{2}\right)}{1}\right]-\frac{1}{4}+1-2\ln \frac{3}{2}$

$=\frac{1}{2}\ln \frac{3}{2}-2\ln \frac{3}{2}+\frac{3}{4}$

$=\ln \frac{3}{2}(\frac{1}{2}-2)+\frac{3}{4}$

$=\frac{3}{4}-\frac{3}{4}\ln \frac{3}{2}$

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

Hence, this is the. Answer.