Thread: Integral questions

Hi, I was wondering if it's possible to find the integrals of these using traditional methods:

$\displaystyle \int x^x dx$

and

$\displaystyle \int \frac{1}{3x^2 + 7} dx$

Another question I have is about definite integrals. When you pull a constant to the left of the integral, do you have to multiply the constant by each function or just to the final answer? For example:

$\displaystyle \int_{0}^{10}{30x}dx$
$\displaystyle 30\int_{0}^{10}{x}dx$
$\displaystyle 30 * \frac{1}{2}x^2 = 15x^2$
$\displaystyle 15(10)^2 - 15(0)^2$

Did I do that right?

Originally Posted by Phire

Hi, I was wondering if it's possible to find the integrals of these using traditional methods:

$\displaystyle \int x^x dx$

I don't think this can be done in terms of elementary functions

and

$\displaystyle \int \frac{1}{3x^2 + 7} dx$

Pull out 1/7 to get

$\displaystyle \frac{1}{7}\int\frac{\,dx}{\frac{3}{7}x^2+1}=\frac {1}{7}\int\frac{\,dx}{\left(\sqrt{\frac{3}{7}}x\ri ght)^2+1}.$

Making the substitution $\displaystyle u=\sqrt{\frac{3}{7}}x$ to transform the integral into

$\displaystyle \frac{1}{\sqrt{21}}\int\frac{\,du}{u^2+1}=\frac{1} {\sqrt{21}}\arctan\!\left(u\right)+C=\frac{1}{\sqr t{21}}\arctan\!\left(\sqrt{\frac{3}{7}}x\right)+C$

Another question I have is about definite integrals. When you pull a constant to the left of the integral, do you have to multiply the constant by each function or just to the final answer? For example:

$\displaystyle \int_{0}^{10}{30x}dx$
$\displaystyle 30\int_{0}^{10}{x}dx$
$\displaystyle 30 * \frac{1}{2}x^2 = 15x^2$
$\displaystyle 15(10)^2 - 15(0)^2$

Did I do that right?

You did it right. When you pull out the constant, it doesn't matter if you multiply it back in once you've integrated of multiply it to the value of the definite integral. You should get the same answer.

Does this make sense? If you have additional questions, please post back.