Answer
0
Work Step by Step
Let \[I=\int_{-3}^{3}\frac{x}{1+|x|}dx\]
For $x<0 , |x|=-x\;\; $ and $\;\;x\geq 0 , |x|=x$
\[I=\int_{-3}^{0}\frac{x}{1-x}dx+\int_{0}^{3}\frac{x}{1+x}dx\;\;\;\ldots (1)\]
Let \[I_1=\int_{-3}^{0}\frac{x}{1-x}dx\]
\[I_1=\int_{-3}^{0}\frac{1-(1-x)}{1-x}dx\]
\[I_1=\int_{-3}^{0}\left(\frac{1}{1-x}-1\right)dx\]
\[I_1=\left[-\ln|1-x|-x\right]_{-3}^{0}\]
\[I_1=-[-\ln 4+3]=\ln 4-3\;\;\;\ldots (2)\]
Let \[I_2=\int_{0}^{3}\frac{x}{1+x}dx\]
\[I_2=\int_{0}^{3}\left(\frac{(1+x)-1}{1+x}\right)dx\]
\[I_2=\int_{0}^{3}\left(1-\frac{1}{1+x}\right)dx\]
\[I_2=\left[x-\ln|1+x|\right]_{0}^{3}\]
\[I_2=3-\ln 4\;\;\;\ldots (3)\]
Using (2), (3) in (1)
\[I=(\ln 4-3)+(3-\ln 4)=0\]
Hence $I=0.$
