Integrate f du/(Gu^2-g) with Step-by-Step Help | Physics Problem

[glueball8]

Homework Statement

How to integrate

f du/(Gu^2-g)

?

Homework Equations

not sure

hmmm thre's the one du/(a^2+b^2)=...

The Attempt at a Solution

But I got a complex number.

This is not HW. hmm just can't solve a physics problem without it
Ps. I never learn integration

 

[Dick]

f, g and G constants? If so either a trig substitution or partial fractions. Depends on signs. Can you be more specific?

 

[avinash patha]

in case g,f and G are constants it can be slvd as
f/G[1/u^2-{(g/G)^1/2}^2] du

F/G ln[{u-(g/G)^1/2}/u+(g/G)^1/2] +C

 

[glueball8]

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv

 

[glueball8]

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv[\tex]<br /> <br /> where p C A m g are constants

 

[Mark44]

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv[\tex]<br /> <br /> where p C A m g are constants

<br /> For some reason your tex stuff isn&#039;t rendering.

 

[Defennder]

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv

where p C A m g are constants

Use this slash instead of the other one to closed tex tags: /tex

 

[glueball8]

Use this slash instead of the other one to closed tex tags: /tex

t=\int_{v_{o}}^{v_{f}} \frac{m}{\frac{1}{2}pC_{d}Av^2-mg}dv


ok thanks.

 

[Mark44]

Factor out all of the stuff that multiplies v^2 in the denominator as well as m in the numerator. Your integral will look like this:
K_1 \int \frac{dv}{v^2 - K_2}
K_1 = 2m/(p*C_d*A) and K_2 = mg/(.5p*C_d*A)

Now, you can do either of two things:
1. factor the denominator and then use partial fraction decomposition to get you antiderivative.
2. look up the resulting integral in, say, the CRC Math Tables.

 

[glueball8]

Let \ \ \frac{1}{2}pCA=G
=m\int_{vo}^{vf}\frac{1}{Gv^2-mg}dv
= m \int_{vo}^{vf}(\frac{\frac{1}{2\sprt{mg}}}{\sqrt{G}v-\sqrt{mg}}+\frac{\frac{-1}{2\sprt{mg}}}{\sqrt{G}v+\sqrt{mg}} dv)
= \frac {m}{\sqrt{mgG}} ln\ ( \ {|} \ \frac{\sqrt{G}v-\sqrt{mg}}{\sqrt{G}v+\sqrt{mg} \ {|} \ } {)} |_{vo}^{vf}

 

[glueball8]

t=\int{v}^{v_{f}} {-} \frac{{1}{}pC_{d}-mg}dv

 

[glueball8]

${\displaystyle Kv_^2 = g - e^-^2^K^y \times (g-Kv_{o}^2)}$