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 Q6   Integrate the functions \sqrt { ax + b }

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Given to integrate \sqrt { ax + b } function,

Let us assume (ax+b) = t

we get, adx =dt

dx = \frac{1}{a}dt

\Rightarrow \int(ax+b)^{\frac{1}{2}} dx = \frac{1}{a}\int t^{\frac{1}{2}}dt

Now, by back substituting the value of t,

= \frac{1}{a}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

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

Posted by

Divya Prakash Singh

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