# Calculus

How would I find the Integral of

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

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

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1. try factoring the bottom and using the triangle rule

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posted by Mischa
2. There's my problem. I just can't factor the bottom.

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posted by Sean
3. Actually, it's arctan.

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

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

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posted by Mischa
4. Oh. 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.

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posted by Sean
5. that 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)

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

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posted by Mischa

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