How to Integrate sqrt(1+3x) Correctly?

sargentidiot
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NVM bout the arc length, need help on integrations

Homework Statement


Integrate sqrt(1+3x)

Homework Equations


Sqrt (1+3x)

The Attempt at a Solution


i made it into
(1+3x)^(1/2)
 
Last edited:
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sargentidiot said:
NVM bout the arc length, need help on integrations

Homework Statement


Integrate sqrt(1+3x)


Homework Equations


Sqrt (1+3x)




The Attempt at a Solution


i made it into
(1+3x)^(1/2)

have you tried u-substitution? use u=1+3x and see what you get
 
ok so i got
(2(1+3x)^(3/2))/9
would that be right
 
sargentidiot said:
ok so i got
(2(1+3x)^(3/2))/9
would that be right

yes, you got it
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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