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What is integration by parts, and how is it useful?

"Integration by parts" is one of several methods in our arsenal that we can use to integrate a function f(x). It involves expressing f(x) as the product of two other functions, which I will call u(x) and v'(x) (where v' = dv/dx is the first derivative of v with respect to x):

 

∫ f(x) dx  =  ∫ u(x· v'(x) dx  =  u(x· v(x)  -  ∫ u'(x· v(x) dx

 

This formula comes from the product rule for differentiation. If we differentiate the product u(x· v(x) with respect to x using the product rule, we get the following:

 

d/dx(u(x· v(x))  =  u(x· v'(x)  +  u'(x· v(x)

 

Integrating both sides of this expression with respect to x removes the d/dx on the left hand side, and creates two integrals on the right hand side:

 

u(x· v(x)  =  ∫ u(x· v'(x) dx  +  ∫ u'(x· v(x) dx

 

If we subtract the final term from both sides, we arrive at our original formula again:

 

∫ f(x) dx  =  ∫ u(x· v'(x) dx  =  u(x· v(x)  -  ∫ u'(x· v(x) dx

 

But why would we want to use this method? Well, integration by parts is generally useful in cases such as the one below:

 

∫ Ln(x)/x2 dx

 

At a first glance, we have no idea how to integrate this function. We do, however, know how to integrate 1/x2 (which is our v' in this case), and we know how to differentiate Ln(x) (which is our u). So, using integration by parts:

 

∫ Ln(x)/x2 dx  =  ∫ Ln(x· (1/x2) dx  =  (-1/x· Ln(x)  -  ∫ (-1/x) · (1/x) dx  

=  -Ln(x)/x  +  ∫ (1/x2) dx  =  -Ln(x)/x  -  1/x  +  C

 

(Not forgetting our constant of integration at the end!)

Alexander J. IB Physics tutor, A Level Physics tutor, GCSE Physics tu...

2 years ago

Answered by Alexander, an IB Maths tutor with MyTutor


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