Question

evaluate the integral dx/sqroot x(1+x)

Answer 1

\[\int\frac{dx}{\sqrt{x(1+x)}}\] ?

Answer 2

si
:)

Answer 3

the sq root
is only for x

Answer 4

decompose it maybe since its already factored?

Answer 5

\[\frac{1}{\sqrt{x}(1+x)} = \frac{A}{\sqrt{x}}+\frac{B}{(1+x)}\]

Answer 6

ok

Answer 7

so what u do?

Answer 8

you solve for A and B by reworking the problem back into the left side;
\[1 = A(1+x) + B(\sqrt{x})\]
when x=-1; 1 = 0 + Bsqrt(x); B = 1/sqrt{x}

when x = 0; 1 = A(1+0) + 0 ; A = 1

Answer 9

that dint work ... missed the sqrt{-1} :)

Answer 10

my integral tables says:

\[\int\frac{dx}{(1+x)\sqrt{x}}=2\ \tan^{-1}\left(\sqrt{x}\right)\]maybe?

Answer 11

tan-1(sqrt(x)) derives to what?

\[\frac{1}{1+\sqrt{x}^2}*\frac{2}{2\sqrt{x}}\]
it works :)

Answer 12

i did not understand

Answer 13

since tan^-1(x) derives down to get 1/(1+x^2) ; then integrals with the same structure will suit back up ....

Answer 14

just as x^3 derives down to 3x^2

3x^2 integrates back up to x^3 +c

Answer 15

the derivate and the integral are inverse functions

Answer 16

I think you make a substitution of u=x^(1/2) and go from there.

Answer 17

aint seen loki in awhile :)

Answer 18

And you've taken over the world in the meantime.

Answer 19

it was either that or .... well there wasnt another option lol