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How do you evaluate the integral #int sqrt(2x+3)dx#?

Answer 1

Christopher Allers

#= 1/3(2x + 3)^(3/2)+ C#

Let #u = 2x + 3#. Then #du = 2dx# and #dx= 1/2du#

#=1/2intsqrt(u)du#

#=1/2int u^(1/2)du#

You can integrate this as #int(x^n)dx = x^(n + 1)/(n + 1) + C#:

#=1/2(2/3u^(3/2)) + C#

#=1/3u^(3/2)+ C#

#= 1/3(2x + 3)^(3/2)+ C#

Hopefully this helps!

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Answer 2

Emilia Aguilar

To evaluate the integral ∫√(2x+3)dx, we use a u-substitution. Let u = 2x + 3. Then, du/dx = 2, and dx = du/2. Substituting these into the integral, we get:

∫√(2x+3)dx = ∫√u * (du/2) = (1/2)∫√u du

Now, we integrate √u with respect to u:

(1/2)∫√u du = (1/2) * (2/3) * u^(3/2) + C = (1/3)u^(3/2) + C

Finally, substituting back u = 2x + 3, we get the result:

(1/3)(2x + 3)^(3/2) + C

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.