U substitution and integration by parts

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robertjford80
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I would think because of this

Typeset text showing that the integral of x dx is x^2/2 + C and the integral of x^2 dx is x^3/3 + C, then that the integral of x times x dx is not the product of the two integrals, and the general rule that the integral of f(x)g(x) dx is not equal to the product of the integrals of f and g.


The following problem:

Handwritten-style integral problem 48: the integral of 3x^5 times the square root of (x^3 + 1), dx.


At this stage they should use integration by parts:

Worked substitution step: u = x^3 + 1 gives du = 3x^2 dx and x^3 = u - 1, so the integral of 3x^5 sqrt(x^3 + 1) dx becomes the integral of (u - 1) sqrt(u) du.


However, maybe integration by parts is only useful when one of the parts is e^x ln or a trigonometric formula.
 
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Integration by parts can be useful whenever the integrand is a product of two functions. But it is not always the easiest method to use. For instance, the integral [itex]\int(u-1)\sqrt{u}du=\int(u^{3/2}-u^{1/2})du[/itex] can easily be solved using the power rule in reverse. Of course, you could solve it using integration by parts as well, but it's just more work than is necessary.
 
What would be your u and what would be your dv?

Using integration by parts might work, but I feel that even if it does work it will be much more work than using u substitution.Is the problem here that you don't find the u substitution they used to be 'legal'?
 
Yea, it doesn't seem legal, because I thought you couldn't take the product of two functions in an integrand.
 
robertjford80 said:
Yea, it doesn't seem legal, because I thought you couldn't take the product of two functions in an integrand.

You need to work on being more precise. The integrand of [itex]\int_0^1 x^2 \mathrm{d}x[/itex] is the product of two functions but is clearly integrable.