Can You Solve This Integral Without Using Integration by Parts?

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SUMMARY

The integral of the function \( \frac{x^3}{\sqrt{x^2 + 1}} \) over the interval [0,1] can be evaluated using integration by parts or substitution. The discussion highlights the integration by parts formula, suggesting to set \( u = x^3 \) and \( dv = (x^2 + 1)^{-1/2} dx \) for simplification. An alternative method proposed is the substitution \( u = x^2 \), which simplifies the integral significantly. Both methods lead to the evaluation of the integral, demonstrating flexibility in approach.

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"Evaluate the integral [0,1] x^3/sqrt[x^2 + 1] by integration by parts"

I know I have to use the integration by parts equation, but I don't know what to make u and what to make dv..
 
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Try rewriting the integral thus;

\int^{1}_{0}\; \frac{x^{3}}{\sqrt{x^2+1}} \; dx = \int^{1}_{0}\; x^3 (x^2 +1)^{-\frac{1}{2}} \;dx

Now, to determine which term to make u and dv, think about which one will simplify your expression most when you differentiated it and set this to u.
 
You could start by rewriting the integrand:

<br /> \frac{x^3}{\sqrt{x^2+1}}=x^3 (x^2+1)^{-1/2} = [x^{-6} (x^2+1)]^{-1/2}<br />

And the apply the integration by parts formula. That should make it easier.
 
OH okay, thank you!
 
Do you have to do it by parts? Much easier to substitute u = x^2
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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