Chipset3600 said:Hello MHB, I'm trying to solve this integral:[TEX]\int x^2arctan(x)dx[/TEX] by parts :
here is a litle bit of my resolution: http://i.imgur.com/dZk8M.jpg
when I tried to solve the integral I named B fell into a sort of lool lool.
Chris L T521 said:All looks good so far. Let's focus on this guy, though:
\[\int\frac{x^3}{1+x^2}\,dx\]
I would first recommend you reduce the integrand using long division. Doing this correctly should yield
\[\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}\]
So, we see that
\[\int\frac{x^3}{1+x^2}\,dx=\int\left(x-\frac{x}{1+x^2}\right)\,dx\]
This should be a straightforward integration; no IBP is needed at this point.
Combine this result with everything else you have and you'll get your answer.
I hope this makes sense!
Fantini said:I have tried another method: use the substitution $u = 1+x^2$. Therefore $du = 2x \, dx$ and $x^2 = u-1$. The integral changes shape into
$$\int \frac{x^3}{1+x^2} \, dx = \frac{1}{2} \int (u-1) \, du.$$
This yields
$$\frac{1}{2} \int (u-1) \, du = \frac{u^2}{4} + \frac{u}{2} + C = \frac{(x^2 +1)^2}{4} + \frac{(x^2 +1)}{2} + C.$$
However, it is a different answer. What is wrong?
The concept of "Help- Integral by Parts" is a method of integration used in calculus to solve integrals that cannot be easily solved by other methods. It involves breaking down a complex integral into simpler parts and applying a formula to solve for the final answer.
The formula for "Help- Integral by Parts" is ∫u dv = uv - ∫v du, where u and v are functions of x and dv and du are their respective derivatives.
In general, it is recommended to choose u as the function that will eventually simplify when differentiated, and dv as the function that will eventually become easier to integrate. However, there is no set rule and it often requires trial and error to determine the best choices for u and dv.
One common mistake is to choose the wrong functions for u and dv, which can lead to a more complex integral instead of simplifying it. It is also important to be careful with signs and constants when integrating by parts, as they can easily be overlooked or misinterpreted.
No, "Help- Integral by Parts" is not applicable to all types of integrals. It is most useful for integrals involving products of functions, and it may not work for integrals with trigonometric or exponential functions. In those cases, other integration techniques should be used.