# How do you find #\int ( 4y + 3) ^ { - \frac { 1} { 2} } d y#?

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To find ∫(4y + 3)^(-1/2) dy, you can use the substitution method. Let u = 4y + 3. Then, du/dy = 4, which implies dy = du/4.

Substituting these into the integral:

∫(4y + 3)^(-1/2) dy = ∫u^(-1/2) * (du/4)

This simplifies to:

(1/4) * ∫u^(-1/2) du

Now, you can integrate u^(-1/2) with respect to u:

∫u^(-1/2) du = ∫u^(1/2) * du

Using the power rule for integration, the integral becomes:

(2/3) * u^(3/2) + C

Substituting back u = 4y + 3:

(2/3) * (4y + 3)^(3/2) + C

So, the integral of (4y + 3)^(-1/2) dy is:

(2/3) * (4y + 3)^(3/2) + C

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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.

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.

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.

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