Asked by Sean
How would I find the Integral of
3dx/(10x^2-20x+20)
I've been stuck on this, and would like an answer ASAP.
3dx/(10x^2-20x+20)
I've been stuck on this, and would like an answer ASAP.
Answers
Answered by
Mischa
try factoring the bottom and using the triangle rule
Answered by
Sean
There's my problem. I just can't factor the bottom.
Answered by
Mischa
Actually, it's arctan.
3/10*integral(1/((x-1)^2+1))dx
=3/10arctan(x-1) + C
3/10*integral(1/((x-1)^2+1))dx
=3/10arctan(x-1) + C
Answered by
Sean
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.
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.
Answered by
Mischa
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
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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