Question

How do I integrate an equation where x is in the numerator and denominator? My equation is x/(sqrt(x^2+6)

Answer 1

thats the derivative of the bottom?

Answer 2

@perl the last number is 6 :)

Answer 3

there are a few things to look at when trying to determine the best process, the first thing i look at is if the top and bottom.

since ln(u) derives to u'/u

Answer 4

$$\large \int \Large \frac{x}{\sqrt{x^2+6}}$$

Answer 5

@amistre64 Could I do a U substitution? or, what should I do next?

Answer 6

spose we wanted to take the deriavtive of: sqrt(x^2+a)

what would it look like?

and yes, a u sub would most likely show us this

Answer 7

Yes, do a u sub

Answer 8

\[\int\limits \frac{ x }{ \sqrt{x^2+6} }dx \implies u = x^2+6, du = 2x dx\] can you finish it off?

Answer 9

or

u = sqrt(x^2 + 6) :)

Answer 10

du = 2x/2sqrt(x^2+6) dx

Answer 11

Yes hehe :)

Answer 12

in some cases we have to decompose into seperate fractions, sometimes we can modify by adding 0 or multiplying by 1, other times its a trig function, other times .... but quite frankly these integrable functions are rare and they are trying to get you comfortable with the process, a false sense of security :)

Answer 13

Haha like almost all polynomial equations have no algebraic solution because almost all polynomials have degree greater than 4 ;p

Answer 14

@Astrophysics Yeah, I can finish it off. Is the answer
1/3(x^2+6)(^3/2) ?

Answer 15

Mhm, not quite, \[\frac{ 1 }{ x^{n} } \implies x^{-n}\]

Answer 16

Oh wow it saved over my work.

Answer 17

oh i see! so u^1/2 or (2x+6)^1/2 is the correct answer?

Answer 18

\[\frac{ du }{ 2 }=xdx \implies \frac{ 1 }{ 2 } \int\limits \frac{ du}{ \sqrt{x} } \implies \frac{ 1 }{ 2 } \int\limits u^{-1/2} du\]

Answer 19

Yes, (2x+6)^(1/2) :)

Answer 20

+C

Answer 21

What! I get a C+? xD
A+ for you, Astrophysics, you're the best. Thank you so much :)

Answer 22

one error
its root u :) @Astrophysics
All the best @Babynini

Answer 23

Yes, nice catch, I was just redoing the work as I saved over my previous latex haha, thanks for noticing :)!

Answer 24

@Babynini Haha, +C after you're done integrating as it's a indefinite integral you must know there is a constant :)

Answer 25

oooh...thanks, forgot about that xP

Answer 26

No worries!