First, I would break this fraction apart. Then simplify the fractions.

integral (4^x)/(2^x) dx + integral (10^x)/(2^x)dx =

int. (4/2)^x + int. (10/2)^x = int(2^x)dx + int(5^x)dx

Now we have the form : integral a^u du = 1/(ln a) *a^u + C

Using the above form for our integration problem, we have : [1/(ln 2)] * 2^x + [1/(ln 5)] * 5^x + C

Simplify. (2^x)/(ln 2) + (5^x)/(ln 5) + C

Check out the link below for a list of integrals

We know that (i) Integral (a^x) = (a^x)/ln(a) +C1 and Integral {u(x)+v(x)}dx= Integral u(x) dx+ Integral v(x) +C2

We use the results of the diffrential calculus to resolve

Integral [[4^x+10^x)/2^x] dx.

= Integral { (2^x)(2^x)+5^x)(2^x)]/2^x}dx

=Integral{(2^x+5^x)*2^x/2^x} dx

=Integral(2^x+5^x)dx, . Now using the results ar (ii),

=Integral(2^x)dx+Integral(5^x) dx

= [2^x/(ln2) +C1] +[5^x/(ln5) + C2], C1 and C2 are constants of integration.

=2^x/(ln2)+5^x/(ln5) + C , where C = C1+C2 is the constant of integration.