Let u = x/(x^2 + a^2)^1/2
and you will find that
(1/a^2)* du
= integral of dx/(x^2+a^2)^3/2
which is the integral you want.
Therefore u/a^2
= (x/a^2)/(x^2 + a^2)^1/2
is the answer.
What substitution could I use to integrate
a/(a^2 + x^2)^3/2 dx
3 answers
Computer program says the answer is
x/(a*(a^2 + x^2)^(1/2))
which is slightly different from your answer. Thanks so much for the help on this one. I was really stuck.
x/(a*(a^2 + x^2)^(1/2))
which is slightly different from your answer. Thanks so much for the help on this one. I was really stuck.
I did x = a tan u
dx = a (sec u)^2 du
int of a/(a^2 + x^2)^3/2 dx
= int of (a sec u)^2/(a^2 + (a tan u)^2)^3/2 du
= int of (a sec u)^2/(a sec u)^3 du
= int of (cos u)/a du
= (sin u)/a + K
since u = atan (x/a)
= x/(a*(a^2 + x^2)^(1/2)) + K
Thanks again...
dx = a (sec u)^2 du
int of a/(a^2 + x^2)^3/2 dx
= int of (a sec u)^2/(a^2 + (a tan u)^2)^3/2 du
= int of (a sec u)^2/(a sec u)^3 du
= int of (cos u)/a du
= (sin u)/a + K
since u = atan (x/a)
= x/(a*(a^2 + x^2)^(1/2)) + K
Thanks again...