# Please check me on this integral!

So I have homework due for calculus. I acknowledge that most of you are smarter than I (for now…), so I humbly come to you seeking answers. I looked through a book of integrals to find this, work through it, and I really don’t think it’s right. Here’the OG problem:

(integral from (-1) to (4)) of (2x)/ ((3x+4)^1/4)) is

8 * ((x/3) - ((4/9)* log (4+3x)))

I get (9.0517). Would anyone care to check me? I’ve done it a few times and come up with the same answer, but I don’t feel very confident about it. Any positive renforcement would be greatly appreciated.

My Mathcad program (Mathsoft Inc.) gives 7.831. The integral equals 2/15*(3x+4)[sup]5/3[/sup]-4/3*(3x+4)[sup]2/3[/sup]

My hand-held calculator (Texas Instruments) concurs with David Simmons on the value (1480/189 = 7.83069) but not the integrated fuction. It gets:

(8/189)(3x+4)[sup]3/4/sup

Odd. When I plug in the limits on your function, David, I get something completely different—6.27968.

To do it by hand, try integrating by parts with

u = 2x

and

dv = dx / [3x+4][sup]1/4[/sup]

I think David had a typo when he was finding the antiderivative (maybe ^(1/3) instead of ^(1/4)?), but both he and Achernar have the definite integral correct. (Achernar’s antiderivative is correct, but that may not be immediately obvious from what you get when you do it by hand).

Good call, Cabbage. When I do the integral by hand, I get (unsimplified):

8 / 63 × (3x + 4)[sup]7/4[/sup] - 32 / 27 × (3x + 4)[sup]3/4[/sup]

However, I don’t think it’s necessary to solve the indefinite integral as stated. I think that the best way to do this problem is with u-substitution, with u = (3x+4)[sup]1/4[/sup]. It really de-uglifies the integrand. The fact that the limits on u work out so well if you do this suggests to me that this is a textbook integral which was intended to be solved with u-substitution. Integration by parts works fine, too, but I personally don’t like integrating fractional powers. And if they’re still teaching u-substitution before integration by parts, moe.ron may not be able to take that route.

As a general hint for solving integrals, one of your best bets is to differentiate your final result, and see if you can make it look like your original integrand. Also, a handy resource for indefinite integrals can be found at integrals.com.

If I put the correct function into Mathcad, I get the above answer so I guess we are all in agreement. I must have entered the wrong function under the integral sign.