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How do you integrate #int root4(x^3)+1dx#?

Answer 1

Christopher Allers

#3/7 x^(7/3)+x+C#

This is #int( (x^3)^(1/4)+1) dx#

#=x^(7/3)/(7/3)+x+C#

#=3/7 x^(7/3)+x+C#

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Answer 2

Paisley Johnson

To integrate ∫(root(4)(x^3) + 1) dx, follow these steps:

Use the substitution method: Let u = x^3. Then du/dx = 3x^2, which implies dx = du / (3x^2).
Substitute u = x^3 and dx = du / (3x^2) into the integral to get ∫(root(4)(u) + 1) * (du / (3x^2)).
Simplify the expression to get ∫((u^(1/4)) + 1) * (du / (3x^2)).
Split the integral into two separate integrals: ∫(u^(1/4)) * (du / (3x^2)) + ∫(1) * (du / (3x^2)).
Integrate each term separately.
- For the first integral, use the power rule for integration: integrate u^(1/4) with respect to u to get (4/5) * u^(5/4).
- For the second integral, integrate 1 with respect to u to get u.
Substitute back u = x^3 into the results obtained in step 5.
Simplify the expressions obtained in step 6.
Your final answer will be the sum of the two simplified expressions.

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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.