Question: Integrals (ax + b)²

Answer:

The anti derivative of (ax + b)² is a function of x whose derivative is (ax + b)².

It is known that,

\begin{align} \frac {d}{dx} ((ax+b)^3) = 3a(ax+b)^2 \end{align}

⇒ \begin{align} (ax + b)^2 =\frac {1}{3a} \frac {d}{dx}(ax+b)^3 \end{align}

∴ \begin{align} (ax + b)^2 = \frac {d}{dx}\left(\frac {1}{3a}(ax + b)^3\right) \end{align}

Therefore, the anti derivative of (ax +b)²

\begin{align} (ax + B)^2 \;is \frac {1}{3a}(ax +b)^3 \end{align}

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