Question

Evaluate the following integral.
$\int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}}dx$

Answer 1

We are asked to find the definite integral. Firstly, we will try to eliminate x present in the numerator. To do so, we take $t={{x}^{2}}+1$ and then change dx as $\dfrac{dt}{2x}$ to eliminate x. So, we are then left with $\int{\dfrac{dt}{2t}}$ and then we will change the limit from x = [2, 4] to a new limit in terms of t. Then, at last, $\int{\dfrac{1}{t}at}=\log t$ and using this, we will find our answer.

Complete step by step answer:
We are asked to integrate a given function between the limit from 2 to 4. The function given to us is $\dfrac{x}{{{x}^{2}}+1}.$
We can see that our denominator is a 2-degree polynomial while the numerator is 1 degree. Also, our denominator has only power 2 of x and a constant. So, we will first use the denominator to cancel the numerator.
Let us take ${{x}^{2}}+1$ as t.
$t={{x}^{2}}+1$
Now, differentiating both the sides, we will get,
$\Rightarrow 2xdx+0=dt$
$\Rightarrow dx=\dfrac{dt}{2x}.....\left( i \right)$
Also, earlier we have the limit x = 2 to x = 4. Now, as $t={{x}^{2}}+1,$ so we have the limit changed as
When x = 2, we have, $t={{2}^{2}}+1=5$
When x = 4, we have, $t={{4}^{2}}+1=17$
So, new limit is t = 5 to t = 17.
Hence, we get,
$\int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\int\limits_{5}^{17}{\dfrac{x}{t}\dfrac{dt}{2x}}$
Cancelling the like terms, we get,
$\Rightarrow \int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\int\limits_{5}^{17}{\dfrac{dt}{2t}}$
Taking the constant outside, we will get,
$\Rightarrow \int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\int\limits_{5}^{17}{\dfrac{dt}{t}}$
We know that,
$\int\limits_{a}^{b}{\dfrac{dx}{x}=\left( \log x \right)_{a}^{b}}$
So, we get,
$\Rightarrow \int\limits_{5}^{17}{\dfrac{dt}{t}}=\left( \log t \right)_{5}^{17}$
$\Rightarrow \int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \left( \log t \right)_{5}^{17} \right]$
Putting the limits, we will get,
$\Rightarrow \int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log 17-\log 5 \right]$
As, $\log a-\log b=\log \left( \dfrac{a}{b} \right),$ we will get,
$\Rightarrow \dfrac{1}{2}\left[ \log \left( \dfrac{17}{5} \right) \right]$

Hence, we have,
$\Rightarrow \int\limits_{2}^{4}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\log \left( \dfrac{17}{5} \right)$

Note: Remember always to change the limit we substitute new things like as we change the limit (x = 2 to x = 4) into (t = 5 to t = 17). We have to present the answer in the simplest form and we should simplify log 17 – log 5 as $\log \left( \dfrac{17}{5} \right)$ using $\log a-\log b=\log \left( \dfrac{a}{b} \right).$