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Answer: $=log\left ( \frac{e}{2} \right )$

Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given: $\int_{0}^{1}\frac{x}{x+1}dx$

Solution:$\int_{0}^{1}\frac{x}{x+1}dx=\int_{0}^{1}\left ( \frac{x+1-1}{x+1} \right )dx$

\begin{aligned} &=\int_{0}^{1}\left(\frac{x+1}{x+1}-\frac{1}{x+1}\right) d x \\ &=\int_{0}^{1}\left(1-\frac{1}{x+1}\right) d x=\int_{0}^{1} 1 d x-\int_{0}^{1} \frac{1}{x+1} d x \end{aligned}

Put $1+x=t\Rightarrow dx=dt$

When$x=0\Rightarrow t=1$ and $x=1\Rightarrow t=2$ in 2nd  integral  .

\begin{aligned} &\int_{0}^{1} \frac{x}{x+1} d x=\int_{0}^{1} 1 d x-\int_{1}^{2} \frac{1}{t} d t \\ &=\left[\frac{x^{0+1}}{0+1}\right]_{0}^{1}-[\log |t|]_{0}^{2} \end{aligned} \quad\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \frac{1}{x} d x=\log |x| \end{array}\right]

\begin{aligned} &=[x]_{0}^{1}-[\log 2-\log 1] \\ &=[1-0]-\log \frac{2}{1} \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a-\log b=\log \frac{a}{b}\right]

\begin{aligned} &=1-\log 2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log e=1] \\ &=\log e-\log 2 & {\left[\log a-\log b=\log \frac{a}{b}\right]} \end{aligned}

$=log\left ( \frac{e}{2} \right )$

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