Question

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

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

Short answer

Ans: The exact value of,${\int }_0^1 \frac{x}{1+x^2}dx=\frac{1}{2}\ln \left|2\right|.$

Step-by-step solution

Step 1. Given information.

given,

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

Step 2. The objective is to determine the exact value of the definite integral.  

Let,

$\begin{array}{r}u=1+x^2\\ du=2xdx\end{array}$

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_0^1 \frac{x}{1+x^2}dx\\ ={\int }_0^1 \frac{1}{2u}du\\ =\frac{1}{2}{\int }_0^1 \frac{1}{u}du\\ =\frac{1}{2}[\ln (\left|u\right|){]}_0^1\\ =\frac{1}{2}\left[\ln \left(\left|1+1^2\right|\right)-\ln \left(\left|1+0^2\right|\right)\right]\\ =\frac{1}{2}(\ln |2|-\ln |1\left|\right)\\ =\frac{1}{2}\ln \left|2\right|\end{array}$

Therefore, the exact value is$\frac{1}{2}\ln \left|2\right|$.

Step 3. Check 

The required graph is,