i need to know how to integrate (int)(2/((e^2x)+4))dx i wud be extremely gr8ful if ne1 cud help! thanx?

Question

the answer is 1/2x - 1/4ln|e^2x + 4| +c but i hav no idea how 2 get there:S

Answer 1

Integral: ∫ [2/{(e^2x)+4}] dx

First consider the expression: 2/{(e^2x) + 4}

= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]

= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]


Hence, ∫[2/{(e^2x) + 4}] dx

= ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx

= ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx

= (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx

Let {e^(2x) + 4} = z. Then dz = 2e^(2x) dx

Hence the given integral becomes:
(1/2) x - (1/4)∫(dz/z)

= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration

Now, substitute the value of 'z' in terms of 'x', thus the given integral becomes
= (1/2) x - (1/4)ln|e^(2x) + 4| + c

Answer 2

Use the substitution e^2x = u, then the integral looks like:
(int) (1/(u^2 + 4u)) du
which is in the tables. After integrating, back substitute and your answer is there.

Answer 3

you have to do a substitution. let u=e^(2x) +4:
du/dx =d(e^(2x) +4)/dx
=d(e^(2x))/dx+d(4)/dx
=e^(2x) *d(2x)/dx +0
=e^(2x) *2d(x)/dx
=e^(2x) *2(1)
=2e^(2x)
=2(e^(2x) +4 -4)
=2(u -4)
dx =du/ 2(u -4)
the integral then becomes (int)(2/u) (du/(2(u -4)):
=(int) 1/(u *(u-4)) du
the rational expression 1/(u *(u-4)) can be broken up into two fractions (the process used to do this is called partial fraction decomposition, but i won't explain it here):
1/(u *(u-4))= -1/(4u) +1/(4(u -4))
the integral is now (int)(-1/(4u) +1/(4(u -4)))du
=-.25*ln|u| +.25*ln|u -4|
=-.25* ln|e^(2x) +4| +.25* ln|e^(2x)|
=.25*2x-1/4ln|e^2x + 4|
=1/2x - 1/4ln|e^2x + 4| +c

Answer 4

You need more help in your grammar than in your math