Integral concept help

[Ian McPherson]

I know that one formula for indefinite integrals states that du/u gives you ln|u| + c. so in the problem integral (3x)/(2+x²), u is 2+x² and du is 2x. So you put a 3/2 on the outside to balance for the 2/3 that you multipy the numerator by to make it the required 2x. i now have (3/2) integral (2x)/(2+x²). Now, since i fit the formula du/u, the answer is (3/2) ln |2+x²| + c.

But the problem proceeding that, integral (x²)/((x³-1)^1/2) is u = x³ -1 and du is 3x². so you have (1/3) integral (3x²)/(x³-1)^1/2. Which is du/u. But when my professor worked it, the answer comes out as 2/3(x³-1)^1/2 + c. My question is not how to work these problems, it's why does the first problem fit du/u and the answer is a natural logarithm, while the second also becomes du/u yet there is no natural logarithm in the answer?

 

[galactus]

Because the latter problem becomes \(\displaystyle \displaystyle\frac{1}{3}\int \frac{1}{\sqrt{u}}du=\frac{2\sqrt{u}}{3}\). See?. It has a radical.

The first one used \(\displaystyle \displaystyle \int\frac{1}{u}du=\ln(u)\)

 

[Ian McPherson]

Oh wow, he never told us about the radical having anything to do with it. Thanks

 

[Deleted member 4993]

Oh wow, he never told us about the radical having anything to do with it. Thanks

He should not have to. You should know:

\(\displaystyle \int x^n dx \ = \ \dfrac{x^{n+1}}{n+1} + C \ \ when \ \ n \ne -1\)

and

\(\displaystyle \int x^{-1} dx \ = \ ln|x| + C \)