x/(x^4+3) and 1/(x^(1/2)-x^(1/4))?

Question

This is a definite integration problem.How to get integrals of
x/(x^4+3) and 1/(x^(1/2)-x^(1/4)).

Answer 1

For the first, you will use the formula \int 1/(a^2+u^2) dx=1/a arctan(u/a)+C:

In this case, u=x^2 so 1/2 du=x dx and a=sqrt(3). Then,

\int x/(x^4+3) dx=
1/2 \int 1/(u^2+3) du=
1/(2 sqrt(3)) arctan(u/sqrt(3))+C=
1/(2 sqrt(3)) arctan(x^2/sqrt(3))+C

For the second, factor x^(1/4) from the denominator and then let u=x^(3/4)-1 so
du=3/4 x^(-1/4) dx or, equivalently, 4/3 du = 1/(x^(1/4)) dx and then use
\int 1/u du=ln |u|+C.

\int 1/(x^(1/2)-x^(1/4))dx=
\int 1/[x^(1/4) (x^(3/4)-1)]dx=
4/3 \int 1/u du=
4/3 ln |u|+C=
4/3 ln|x^(3/4)-1|+C

Answer 2

I know 2+2=4