Question

int x/sqrt(x+1)dx please assist :)

Answer 1

sorry, missed the sqrt part. just a sec.

Answer 2

i think this is a trig substitution and i'm so bad at those lol

Answer 3

try x+1 = u^2

Answer 4

\[\int\limits \frac{ x }{ \sqrt{x+1} }dx=\int\limits x \left( x+1 \right)^{-\frac{ 1 }{ 2 }}\,dx\]
you could do integration by parts... have you learned that yet?

Answer 5

yeah, would it be u=(x+1)^(1/2), dv=xdx

Answer 6

no, let u = x

Answer 7

then du = dx and you get rid of the "bad" part. you can integrate 1/(x+1)

Answer 8

you missed the sqrt again lol

Answer 9

sorry, that doesn't prevent it from being integrable, though

Answer 10

ok so do you end up with

Answer 11

u/sqrt(x+1) du

Answer 12

\[\int\limits \frac{ dx }{ \sqrt{x+1} }=\int\limits u^{-\frac{ 1 }{ 2 }}du=2u^{\frac{ 1 }{ 2 }}+c\]

Answer 13

\[=2\sqrt{x+1}\]

Answer 14

+c

Answer 15

but you just lost the xdx and replaced it with just dx
how did that happen