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Q4  Find an anti derivative (or integral) of the following functions by the method of inspection. ( ax + b )^2

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GIven ( ax + b )^2;

So, the anti derivative of ( ax + b )^2 is a function of x whose derivative is ( ax + b )^2.

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

\Rightarrow (ax+b)^2 =\frac{1}{3a}\frac{d}{dx}(ax+b)^3

\therefore (ax+b)^2 = \frac{d}{dx}[\frac{1}{3a}(ax+b)^3]

Therefore, we have the anti derivative of (ax+b)^2 is  [\frac{1}{3a}(ax+b)^3].

Posted by

Divya Prakash Singh

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