Question

How do you find the definite integral of $\int{\dfrac{{{x}^{2}}-2}{x+1}dx}$ from $\left[ 0,2 \right]$?

Answer 1

Hint: We first discuss the integration of functions in binary operations. We take integration of the functions separately with respect to $x$ an apply the same operation on them. We take the operation’s answer as the final solution. We apply the ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for our integration of the numerator ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$. We use the formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$, $\int{\dfrac{1}{x}dx}=\log \left| x \right|+c$.

Complete step by step solution:
We need to find the definite integral of $\int{\dfrac{{{x}^{2}}-2}{x+1}dx}$ from $\left[ 0,2 \right]$. Let $I=\int\limits_{0}^{2}{\dfrac{{{x}^{2}}-2}{x+1}dx}$.
Now we express the numerator of the faction ${{x}^{2}}-2$ as ${{x}^{2}}-2={{x}^{2}}-1-1$.
Now we break the total function into two sub functions.
We get $\dfrac{{{x}^{2}}-2}{x+1}=\dfrac{{{x}^{2}}-1-1}{x+1}=\dfrac{{{x}^{2}}-1}{x+1}-\dfrac{1}{x+1}$.
We apply the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to get ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$.
From the fraction we get $\dfrac{{{x}^{2}}-1}{x+1}=\dfrac{\left( x+1 \right)\left( x-1 \right)}{\left( x+1 \right)}=\left( x-1 \right)$.
So, $I=\int\limits_{0}^{2}{\left( x-1 \right)dx}-\int\limits_{0}^{2}{\dfrac{1}{x+1}dx}$.
We know $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ and $\int{\dfrac{1}{x}dx}=\log \left| x \right|+c$.
We can also form the integration $\int\limits_{0}^{2}{\dfrac{1}{x+1}dx}$ as $\int\limits_{0}^{2}{\dfrac{1}{x+1}d\left( x+1 \right)}$.
This is possible because of $d\left( x+1 \right)=dx$.
Therefore,
$\begin{align}
  & I=\int\limits_{0}^{2}{\left( x-1 \right)dx}-\int\limits_{0}^{2}{\dfrac{1}{x+1}dx} \\
 & =\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{0}^{2}-\left[ \log \left| x+1 \right| \right]_{0}^{2} \\
\end{align}$
We put the values in the equations to get
$I=\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{0}^{2}-\left[ \log \left| x+1 \right| \right]_{0}^{2}=\left[ 2-2 \right]-\left[ \log 3-\log 1 \right]=-\log 3$
Therefore, the definite integral of $\int{\dfrac{{{x}^{2}}-2}{x+1}dx}$ from $\left[ 0,2 \right]$ is equal to $-\log 3$.

Note: We can also solve the second integration using the base change for ratio $z=x+1$. In that case the sum gets complicated but the final solution would be same. It is better to watch out for the odd power value in the ratios and take that as the change in the variable.