Question

What is the antiderivative of, $\left( x+1 \right)\left( 2x-1 \right)$ ?

Answer 1

Hint: The anti-derivative of an expression is the integral of the expression over a certain variable. Now, the expression given to us in the problem is in factored form. So, at first we will break this expression into its individual parts and then apply the formula of integration of a function raised to some constant. This is given as: $\int{{{x }^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\text{ }\left( \text{where, n}\ne -1 \right)$

Complete step by step solution:
The expression given to us in our problem is equal to: $\left( x+1 \right)\left( 2x-1 \right)$. Now,
Let us first assign some terms that we are going to use later in our problem.
Let the given term on which we need to operate an integral be given by ‘y’ . Here, ‘y’ is given to us as:
$\Rightarrow y=\left( x+1 \right)\left( 2x-1 \right)$
The expression written above can be simplified into its individual parts as follows:
$\begin{align}
&\Rightarrow y=2{{x}^{2}}-x+2x-1 \\
&\Rightarrow y=2{{x}^{2}}+x-1 \\
\end{align}$
Then, we need to find the integral of ‘y’ with respect to very small changes in ‘x’. This can be
done as follows:
$\begin{align}
&\Rightarrow \int{y.dx}=\int{\left( 2{{x}^{2}}+x-1 \right)dx} \\
&\Rightarrow \int{y.dx}=2\int{{{x}^{2}}dx}+\int{xdx}-\int{dx} \\
\end{align}$
Now, applying the formula of integration of a function raised to some constant, that is equal to:
$\Rightarrow \int{{{x }^{n}}dx}=\dfrac{{{x }^{n+1}}}{n+1}\text{ }\left( \text{where, n}\ne -1
\right)$
Our expression can be further simplified to:
$\begin{align}
&\Rightarrow \int{y.dx}=\int{2{{x}^{2}}dx}+\int{xdx}-\int{dx} \\
&\Rightarrow \int{y.dx}=2\left[ \dfrac{{{x}^{2+1}}}{2+1} \right]+\left[ \dfrac{{{x}^{1+1}}}{1+1}
\right]-\left[ \dfrac{{{x}^{0+1}}}{0+1} \right]+c \\
&\therefore \int{y.dx}=\dfrac{2{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2}-x+c \\
\end{align}$
Here, the additional term ‘c’ added in our equation is the constant of integration which can take any arbitrary value, as long as nothing is specified in the problem.
Thus, our final expression comes out to be $\dfrac{2{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2}-x+c$ .
Hence, the anti-derivative of, $\left( x+1 \right)\left( 2x-1 \right)$ comes out to be
$\dfrac{2{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2}-x+c$.

Note: While performing an integral, we should make sure that all the formulas are correctly applied. Also, the formula: $\int{{{x }^{n}}dx}=\dfrac{{{x }^{n+1}}}{n+1}\text{ }\left( \text{where,n}\ne -1 \right)$ is valid only when ‘n’ is a constant. That is we cannot apply this formula when the given power is another function, as chain rule has no meaning in integration.