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Find the antiderivative of ${{\left( ax+b \right)}^{2}}$ with respect to x.

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Last updated date: 16th Jul 2024
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Answer
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Hint: To find the antiderivative of ${{\left( ax+b \right)}^{2}}$ with respect to x, we have to integrate ${{\left( ax+b \right)}^{2}}$ with respect to x. For this, we will substitute $t=\left( ax+b \right)$ and differentiate this with respect to x to find the value of dx. Then, we will substitute these values in $\int{{{\left( ax+b \right)}^{2}}dx}$ and integrate with respect to t. Finally, we have to substitute back for t.

Complete step by step answer:
We have to find the antiderivative of ${{\left( ax+b \right)}^{2}}$ with respect to x. We know that anti-derivative means integral. Therefore, we have to find the integral of ${{\left( ax+b \right)}^{2}}$ .
$\Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}...\left( i \right)$
Let us substitute $t=\left( ax+b \right)...\left( ii \right)$ .
We have to differentiate (ii) with respect to x. We know that $\dfrac{d}{dx}a{{x}^{n}}=a\dfrac{d}{dx}{{x}^{n}},\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\text{ and }\dfrac{d}{dx}\left( \text{constant} \right)=0$ . Therefore, we can write the derivative of equation (ii) as
$\begin{align}
  & \Rightarrow \dfrac{dt}{dx}=a\times 1+0 \\
 & \Rightarrow \dfrac{dt}{dx}=a \\
 & \Rightarrow \dfrac{dt}{a}=dx...\left( iii \right) \\
\end{align}$
Let us substitute (ii) and (iii) in (i).
$\Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\int{{{t}^{2}}\dfrac{dt}{a}}$
Let us take the constant outside. We can write the above equation as
$\Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\dfrac{1}{a}\int{{{t}^{2}}dt}$
We know that \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C\] . Therefore, we can write the above integral as follows.
\[\begin{align}
  & \Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\dfrac{1}{a}\left( \dfrac{{{t}^{2+1}}}{2+1} \right)+C \\
 & \Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\dfrac{1}{a}\left( \dfrac{{{t}^{3}}}{3} \right)+C \\
 & \Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\dfrac{1}{3a}{{t}^{3}}+C \\
\end{align}\]
Let us substitute the value of t from equation (ii) in the above equation.
$\Rightarrow \int{{{\left( ax+b \right)}^{2}}dx}=\dfrac{1}{3a}{{\left( ax+b \right)}^{3}}+C$
Hence, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\dfrac{1}{3a}{{\left( ax+b \right)}^{3}}+C$ .

Note: Students must be thorough with integration and differentiation of basic functions, how to integrate and differentiate functions and the rules associated with these. They must never forget to add the constant, C after the integration of functions. They must never forget to substitute back for t. Students must know that antiderivative is also known as inverse derivative, primitive function, primitive integral or indefinite integral.