Question

Find an anti derivative (or integral) of function by the method of inspection
(ax + b)².​

Answer 1

Answer:

We know that,

dxd(ax+b)3=3a(ax+b)2

⇒(ax+b)2=3a1dxd(ax+b)3

∴(ax+b)2=dxd(3a1(ax+b)3)

Therefore, the anti derivative of (ax+b)2 is  3a1(ax+b)3.

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Answer 2

Question:-

Find an anti derivative (or integral) of function by the method of inspection (ax + b)².

Given:-

• An Integral (ax + b)².

To Find:-

An anti derivative (or integral) of function by the method of inspection (ax + b)².

Solution:-

Let, f(x) = (ax + b)²

Note that, $\sf \large \frac{d((ax + b) {}^{3} )}{dx} = 3a(ax + b) {}^{2}$

$\sf \large \therefore (ax + b) {}^{2} = \frac{1}{3a} \: \: \frac{d((ax + b) {}^{3} )}{dx}$

$\sf \large= \frac{d}{dx} \bigg ( \frac{1}{3} (ax + b {}^{3} )$

Answer:-

$\sf \large \color{red}Hence, the \: anti - derivative \: of \: (ax + b) {}^{2} \: is \: \frac{1}{3a} (ax + b) {}^{3} .$

Hope you have satisfied. ⚘