∫_(-1)^1 {dx/ (t ^(2)-25)}

you dont have to solve i just want to know whats my u

Question

∫_(-1)^1 {dx/ (t ^(2)-25)}

you dont have to solve i just want to know whats my u

amistre64 · Answer

hyperbolic ?

anonymous · Answer

I suppose technically there is no "u" because it is slightly more complicated than a u-sub. You would have to do trig sub if I'm seeing it right.

amistre64 · Answer

can you z sub it?  like you gotta use a z lol

anonymous · Answer

$$\int\limits_{0}^{4}{\frac{2t}{t^2-25}}dt$$ sorry this is the corect question

amistre64 · Answer

t = z^(1/2) or so?

anonymous · Answer

u sub

amistre64 · Answer

thats just an ln

amistre64 · Answer

the top is the derivative of the bottom; it been ln-ized

anonymous · Answer

ok...i dont understand what your saying... amistre

amistre64 · Answer

$$\int\frac{dx}{x}\implies ln|x|+c$$

amistre64 · Answer

in this case; x = t^2-25; and dx = 2t

anonymous · Answer

1/x yes....but i dont see it in that integral

anonymous · Answer

Yeah, you can use u=t^2-25. Then du=2t
Then your integral becomes
$$\int\limits \frac{du}{u}$$
You need to re-evaluate your limits of integration however.

amistre64 · Answer

whenever the top is the derivative of the bottom; you got i log intergral

anonymous · Answer

we didnt do limits of integration

amistre64 · Answer

no need to go thru the extra work; ln(t^2-25) is your integral

amistre64 · Answer

[ln(x^2)]' = 2x/x^2 ; the top is the derivative of the bottom thanks to the chain rule

anonymous · Answer

You have the integral from 0 to 4. Those are your limits.

anonymous · Answer

ok

amistre64 · Answer

ln|t^2-25| then ;)

anonymous · Answer

okay thank you both

anonymous · Answer

ok i dont see ow to get the answe

amistre64 · Answer

the answer is just in recognizing the integral as an adaptation to $\int\frac{dx}{x}$

anonymous · Answer

then?

amistre64 · Answer

then you subtract the ln|t^2 -25| using the upper and lower bounds

F(upper) - F(lower)

amistre64 · Answer

$$ln|(4^2)-25|-ln((0)^2-25)=?$$

amistre64 · Answer

there spose to be absolute value signs around the ln parts

amistre64 · Answer

ln(9) - ln(25) = ?

amistre64 · Answer

ln(9/25) = ?

anonymous · Answer

i thought u cant find the ln of a negative number

amistre64 · Answer

you cant; which is why you absolute value it to make it positive

anonymous · Answer

oh ok

amistre64 · Answer

the books show the correct way to do it is with: $$ln|x|$$ and then give a reason why its good either way, but i cant recall the details :)

anonymous · Answer

okay thank you wow i understand now!!!

amistre64 · Answer

:) youre welcome