How would I solve the following integral dx / (5+x^2) ?

Answer 1

does f'(x) = 5+x^2?

if so:

let s stand for "integral of"

f'(x) = 5+x^2

f(x) = s(5+x^2)dx

f(x) = s5dx + sx^2dx

f(x) = 5x + (1/3)x^3

Answer 2

Smells like an arc tangent.

Substitute

x = (sqrt 5) tan u, then
u = arc tan [x / sqrt(5)]
dx = (sqrt 5) [(tan u)^2 + 1] du

5 + x^2 = 5 [(tan u)^2 + 1]

The integral becomes

INT du / sqrt 5 = u / sqrt 5

= {arc tan [x / sqrt 5]} / sqrt 5

Answer 3

Use the following substitution:
Let x=sqrt(5)tan(y).

And then you should be able to do the rest.

Answer 4

yup, it's an arctg.. or arctan.. whatever it's acalled in Egnlish... and somewhere u have the formula 1/(a^2+x^2)=arctgx/a or something like that...u check up on it, i just gave u the hint... can't do all the work..