Question

integral question below

Answer 1

\[\int\limits_{}^{}\frac{ x }{ \sqrt{x ^{2}+4x+3}}dx =\int\limits \frac{ x }{\sqrt{ (x+2)^{2}-1}}\]

Answer 2

i got a little bit more work so far hold on

Answer 3

x+2 = sec(x) is what im told to do.. and i get this then
x= sec(x)-2

Answer 4

i would have written
x = A d/dx (x^2+4x+3) + B
find A,B , then separate the denominator....
but i still have to figure out the best way....

Answer 5

dx=sec(beta)tan(beta)d(beta)

^ yes i forgot to change betas and x symbols but .... ya kno what i meant XD

Answer 6

if you are told to substitute that , then did u get
(sec -2) sec ?

Answer 7

\[\int\limits_{}^{}\frac{ \sec(u)-2 }{ \sqrt{\sec(u)^{2}}-1 }\sec(u)\tan(u)du\]

Answer 8

the -1 is inside the sqrt sign
and sec^2 -1 =... ?