Integral of (x^2)(sqrt(x^3 + 1))dx. Please check my work? Thank you

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SUMMARY

The integral of the function ∫x^2(√(x^3 + 1))dx was evaluated using the substitution method. The substitution u = x^3 + 1 leads to du = 3x^2dx, allowing the integral to be rewritten as (1/3)∫√(u)du. The final result is 2/9(x^3 + 1)^(3/2) plus a constant of integration, which is crucial to include. The discussion emphasizes the importance of differentiating the result to verify correctness.

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Lo.Lee.Ta.
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∫x^2(√(x^3 + 1))


u = x^3 + 1
du = 3x^2dx
(1/3)du = x^2dx

∫√(u)*(1/3)du

= 2/3(u)^3/2 *(1/3)

= 2/9(x^3 + 1)^3/2

Is this the correct answer? Thank you.
 
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Yep, looks like it. You can always differentiate your answer if you're not sure.
 
Lo.Lee.Ta. said:
∫x^2(√(x^3 + 1))


u = x^3 + 1
du = 3x^2dx
(1/3)du = x^2dx

∫√(u)*(1/3)du

= 2/3(u)^3/2 *(1/3)

= 2/9(x^3 + 1)^3/2

Is this the correct answer? Thank you.

Yes, but don't forget about the constant of integration.
 
Thanks, you guys! :)

That's true. I should differentiate when I'm not sure...
 
Lo.Lee.Ta. said:
Thanks, you guys! :)

That's true. I should differentiate when I'm not sure...

You should differentiate even when you are sure! (Sometimes a person can be sure but still have the wrong answer.)

In other words: always differentiate!
 

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