Question

The value of the definite integral ${\int }_{-1}^{1}xIn(1^x+2^x+3^x+6^x)dx$ equals

• A. $\frac{In2+In3}{2}$
• B. $\frac{In2+In3}{3}$
• C. $In2+In3$
• D. $\frac{In2+In3}{4}$

Answer 1

Answer: (B)

$\frac{In2+In3}{3}$

$I={\int }_{-1}^{1}x\ln (1^x+2^x+3^x+2^x{.3}^x)dx$

$I={\int }_{-1}^{1}x\ln \left[\right(1+2^x).(1+3^x\left)\right]dx$

$I={\int }_{-1}^{1}x\ln (1+2^x)dx+{\int }_{-1}^{1}x\ln (1+3^x)dx$

$I_1={\int }_{-1}^{1}x\ln (1+2^x)dx....\left(1\right)$

Replace $x$ by $a+b-x\Rightarrow -x$

$\Rightarrow I_1{\int }_{-1}^1-x\ln \left(\frac{1+2^x}{2^x}\right)dx={\int }_{-1}^1-x\ln (1+2^x)dx={\int }_{-1}^1-x\ln (1+2^x)dx+{\int }_{-1}^{1}x\ln 2^{x}dx....\left(2\right)$

Add $\left(1\right)$ and $\left(2\right)$

$2I_1=\ln {\int }_{-1}^1{x}^{2}dx$

$I_1=\frac{\ln 2}{2}{\left[\frac{x^2}{3}\right]}_{-1}^1=\frac{\ln 2}{2}\times \frac{2}{3}=\frac{\ln 2}{3}$

Similarly, $I_2=\frac{\ln 3}{3}$

Thus $I=I_1+I_2$

$=\frac{\ln 2+\ln 3}{3}\left(B\right)$