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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asked by Sean
  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)

    so your answer is:
    4*arcsin((t+1)/4) + C

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    posted by Mischa

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