Question

Evaluate the following definite integral:
$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}}dx$

Answer 1

Hint: To evaluate the above integral, we have to first find the indefinite integration of $\dfrac{x}{{{x}^{2}}+1}$ and then apply the lower and upper limit on the indefinite integration. Now, to find the indefinite integration, let us assume ${{x}^{2}}+1$ as “t” then take the differentiation on both the sides and after the differentiation, you will find that the numerator which is given in the above problem is the differentiation of the denominator then substitute “t” in place of x in the given integral and also change the limits of the given integral because now, we are integrating with respect to “t”.

Complete step-by-step solution:
The definite integral given in the above problem is equal to:
$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}}dx$
Let us assume ${{x}^{2}}+1$ given in the above integral as “t”.
${{x}^{2}}+1=t$ ……… Eq. (1)
Now, differentiating on both the sides we get,
$\begin{align}
  & 2xdx+0=dt \\
 & \Rightarrow 2xdx=dt.......Eq.(2) \\
\end{align}$
Now, replace all the x in the given integral to t so the limits will also change.
The lower limit of the new integral is equal to:
${{x}^{2}}+1=t$
Substituting the value of x as 0 which is the lower limit of the integral in x we get the lower limit of integration in “t” we get,
$\begin{align}
  & 0+1=t \\
 & \Rightarrow t=1 \\
\end{align}$
Now, substituting the value of x as 1 in ${{x}^{2}}+1=t$ we get the upper limit of “t”.
$\begin{align}
  & {{1}^{2}}+1=t \\
 & \Rightarrow 2=t \\
\end{align}$
Hence, we got the lower and upper limit of the integral in “t” is 1 and 2 respectively.
Now, replacing the integral in x to integral in “t” we get,
$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}}dx$
Substituting ${{x}^{2}}+1=t$, lower and upper limit as 1 and 2 respectively we get,
$\int\limits_{1}^{2}{\dfrac{x}{t}}dx$
From eq. (2) we get,
$2xdx=dt$
Dividing 2 on both the sides we get,
$xdx=\dfrac{dt}{2}$
Substituting the above value of $xdx$ in $\int\limits_{1}^{2}{\dfrac{x}{t}}dx$ we get,
$\int\limits_{1}^{2}{\dfrac{1}{2t}}dt$
Taking constant $\dfrac{1}{2}$ out from the above integral we get,
$\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{1}{t}}dt$
We know that, the indefinite integral of $\dfrac{1}{x}$ is equal to:
$\int{\dfrac{1}{x}}dx=\ln x+c$
Using the above integration in evaluating the definite integral in “t” we get,
$\dfrac{1}{2}\left| \ln t+c \right|_{1}^{2}$
Applying the lower and upper limits in the above we get,
$\begin{align}
  & \dfrac{1}{2}\left( \ln 2+c-\ln 1-c \right) \\
 & =\dfrac{1}{2}\left( \ln 2-\ln 1 \right) \\
\end{align}$
We are going to use the following property of logarithm to solve the above expression:
$\ln a-\ln b=\ln \dfrac{a}{b}$
$\begin{align}
  & \dfrac{1}{2}\ln \left( \dfrac{2}{1} \right) \\
 & =\dfrac{1}{2}\ln 2 \\
\end{align}$
Hence, we got the value of the given definite integral as $\dfrac{1}{2}\ln 2$.

Note: You can, even more, simplify the final evaluation of the definite integral that we are getting above.
The final value of the given integral that we are getting above is:
$\dfrac{1}{2}\ln 2$
You can write the above answer in the more simplified form by writing the value of ln2. The value of ln2 is equal to 0.693 so substituting the value of ln2 as 0.693 in the above we get,
$\begin{align}
  & \dfrac{1}{2}\left( 0.693 \right) \\
 & =0.3465 \\
\end{align}$
It is not necessary to write the simplified form of $\dfrac{1}{2}\ln 2$ because sometimes in the multiple-choice questions, the options are in this form only. You can show this answer in the most simplified form in the subjective exam or if in the objective exam, options contain this simplified form.