comAttitRize8

2022-07-20

Find the antiderivative of $\sqrt{3x-1}dx$.
I got $\frac{2}{3}\left(3x-1{\right)}^{3/2}+c$ but my book is saying $\frac{2}{9}\left(3x-1{\right)}^{3/2}+c$.
Can some one please tell me where the 2/9 comes from?

Reese King

Expert

Step 1

Let $u=3x-1$, , so

Add one to the power of u, and divide by the new power

Step 2
Your problem is that you need to take into consideration the derivative of Your problem is that you need to take into consideration the derivative of $3x-1$, which means you need to divide by 3, giving you the factor of $\frac{1}{3}$ missing from your answer.
As was suggested in the comments of your previous question, if you differentiate the answer that you got with the chain rule, then you'll see why you've made a mistake
$\frac{d}{dx}\left(\frac{2}{3}\left(3x-1{\right)}^{3/2}\right)=\frac{3}{2}\cdot \frac{2}{3}\left(3x-1{\right)}^{1/2}\underset{=3}{\underset{⏟}{\frac{d}{dx}\left(3x-1\right)}}=3\sqrt{3x-1}$

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