How would I find the Integral of

Question

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

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

Answer 1

try factoring the bottom and using the triangle rule

Answer 2

There's my problem. I just can't factor the bottom.

Answer 3

Actually, it's arctan.

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

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

Answer 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.

Answer 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