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July 31, 2014

July 31, 2014

Posted by **Sean** on Wednesday, February 13, 2008 at 10:17pm.

3dx/(10x^2-20x+20)

I've been stuck on this, and would like an answer ASAP.

- Calculus -
**Mischa**, Wednesday, February 13, 2008 at 10:32pmtry factoring the bottom and using the triangle rule

- Calculus -
**Sean**, Wednesday, February 13, 2008 at 10:34pmThere's my problem. I just can't factor the bottom.

- Calculus -
**Mischa**, Wednesday, February 13, 2008 at 10:38pmActually, it's arctan.

3/10*integral(1/((x-1)^2+1))dx

=3/10arctan(x-1) + C

- Calculus -
- Calculus -
**Sean**, Wednesday, February 13, 2008 at 10:45pmOh. Okay thanks.

Thank you so much. If it's not to much...

How would I integrate

(4dt)/(sqrt(15-2t-t^2))

That thing has me stumped.

- Calculus -
**Mischa**, Wednesday, February 13, 2008 at 11:36pmthat one's arcsin.

factor it into:

4*integral(1/sqrt(16-(t^2+2t+1)))dt

so,

4*integral(1/sqrt(4^2+(t+1)^2))dt

divide everything in the sqrt by 4^2 and bring it out as 1/4,

4/4*integral(1/sqrt(1+((t+1)/4)^s))dt)

now it's in arcsin form (remember u and du)

so your answer is:

4*arcsin((t+1)/4) + C

- Calculus -

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