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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\begin{align} \frac{3}{2}(2x+1)\end{align}
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\begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align}