Solving this definite integral using integration by parts

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  • #1
songoku
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Homework Statement
Please see below
Relevant Equations
Integration
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Using integration by parts:
$$I_n=\left. x(1+x^2)^{-n} \right|_0^1+\int_0^{1} 2nx^2(1+x^2)^{-(n+1)}dx$$
$$I_n=2^{-n} + 2n \int_0^{1} x^2(1+x^2)^{-(n+1)}dx$$
Then how to continue?

Thanks
 
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  • #2
My first approach would be work out what In+1 is and see how they relate.

Their hint of multiplying by 1 tells me that I might need to represent 1 as a fraction where numerator = denominator.
 
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  • #3
Hint: ##x^2 = (x^2+1) - 1##.
 
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  • #4
I understand

Thank you very much scottdave and vela
 
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FAQ: Solving this definite integral using integration by parts

What is integration by parts?

Integration by parts is a technique used to integrate the product of two functions. It is based on the product rule for differentiation and is given by the formula: ∫u dv = uv - ∫v du, where u and dv are differentiable functions of x.

How do I choose u and dv in integration by parts?

Choosing u and dv can be tricky, but a common heuristic is the LIATE rule, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. Generally, you choose u to be the function that comes first in this list and dv to be the remaining part of the integrand.

What do I do if integration by parts needs to be applied more than once?

If applying integration by parts once does not simplify the integral sufficiently, you may need to apply it multiple times. After each application, you should simplify the resulting integrals and check if further applications are necessary.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for definite integrals. The formula is modified to account for the limits of integration: ∫[a,b] u dv = [uv] from a to b - ∫[a,b] v du. You compute uv at the bounds and then subtract the integral of v du over the same interval.

What if the integral cycles back to the original integral?

If the integral cycles back to the original integral after applying integration by parts, you can solve for the original integral algebraically. Set up an equation where the original integral appears on both sides, then solve for the integral by isolating it on one side of the equation.

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