Which term should be chosen for u(x) in integration by parts?

[zanazzi78]

ok i`m really struggling with the concept.
I've been asked to find the indefinite integral of;

[tex] \int \frac{x^2}{(2+ x^3)} dx [/tex]

so before i beg for the answer could someone confirm that i`ve got the right rule to solve this;

[tex] \int u(x) v'(x) = [ u(x) v(x)] - \int v(x) u'(x) [/tex]

if this is right would you mind giving a suggestion to what u(x) to use?

p.s. i may have to edit this if latex doesn`t come out right I've been having trouble with it and only jointed the forum a few day's ago!

 

[HallsofIvy]

Yes, that is the correct formula (it's surely in your textbook), but I don't see any good reason for using integration by parts. That's a fairly standard anti-derivative: look up "arctangent".

 

[zanazzi78]

I`ve just edited it. Now it`s integration by parts! see I`m getting all flustered

 

[Data]

Actually, you still don't need integration by parts. Substituting [itex]u = x^3[/itex] will simplify it to a standard form.

Anyways, since differentiation is in some sense the inverse operation to integration, every differentiation rule yields an integration rule. I'm sure you remember the product rule:

[tex] \frac{d}{dx}(f(x) g(x)) = f^\prime (x) g(x) + g^\prime (x) f(x)[/tex]

Rearraging the equation above gives

[tex] f(x) g^\prime (x) = \left[f(x) g(x)\right]^\prime - f^\prime(x)g(x)[/tex]

And integrating both sides with respect to x gives

[tex] \int f(x) g^\prime (x) \ dx = \int \left(\frac{d}{dx} (f(x) g(x)) - f^\prime(x)g(x)\right) \ dx[/tex]

but certainly, [tex] \int \frac{d}{dx}(f(x)g(x)) \ dx = f(x)g(x)[/tex], so this just reduces to

[tex] \int f(x)g^\prime (x) \ dx = f(x)g(x) - \int g(x) f^\prime (x) \ dx[/tex]

which is what your teacher probably called the formula for integration by parts.

 

[dextercioby]

[tex] x^{2}dx=\frac{1}{3}d(2+x^{3}) [/tex] so the integration is simple...

Daniel.

 

[BobG]

ok i`m really struggling with the concept.
I've been asked to find the indefinite integral of;

[tex] \int \frac{x^2}{(2+ x^3)} dx [/tex]

so before i beg for the answer could someone confirm that i`ve got the right rule to solve this;

[tex] \int u(x) v'(x) = [ u(x) v(x)] - \int v(x) u'(x) [/tex]

if this is right would you mind giving a suggestion to what u(x) to use?

p.s. i may have to edit this if latex doesn`t come out right I've been having trouble with it and only jointed the forum a few day's ago!

As others have said, you don't have to use integration by parts to solve this, but ...

Pick the term whose derivative will eventually go to zero (the soonest). In this case, it's x^2. First derivative is 2x. Second derivative is 2. Third derivative is 0.