Integrals on exam I couldn't answer

[noboost4you]

On a recent exam I took, these integrals came up and I was unable to answer them correctly:

<br /> \int \frac{2}{(x-1)^2+1} dx<br />

and

<br /> \int \frac{2}{\sqrt{1-(x-1)^2}} dx<br />

I had absolutely no idea how to solve those 2 equations and my professor decided not to go over any of the problems from the test. He did mention that I needed to use substitution, but I am still unclear of how to solve them.

Can anyone help me out?

Thanks

 

[noboost4you]

Originally posted by Ambitwistor
Do you know how to do these integrals?

<br /> \int \frac{1}{x^2+1}\,dx<br />

and

<br /> \int \frac{1}{\sqrt{1-x^2}}\,dx<br />

<br /> \int \frac{1}{\sqrt{1-x^2}}\,dx<br /> = inverse sine

and

<br /> \int \frac{1}{x^2+1}\,dx<br /> = inverse tangent

I only know those answers by rule, when he added more constants and suggested substitution is when I didnt understand.

 

[noboost4you]

<br /> \int \frac{2}{(x-1)^2+1} dx<br /> would equal

2\int \frac{1}{(x-1)^2+1} dx
then u = x-1
du = dx
the new integral then equals
2\int \frac{1}{u^2 +1} du
which in turn equals 2tan^-1(x-1) ??

and

<br /> \int \frac{2}{\sqrt{1-(x-1)^2}} dx<br /> would equal

2\int \frac{1}{\sqrt{1-(x-1)^2}} dx
then u = x-1
du = dx
the new integral then equals
2\int \frac{1}{\sqrt{1-u^2}} du
which in turn equals 2sin^-1(x-1) ??

am I correct?

Thanks again

 

[noboost4you]

Great! Only if I would have known that last week ;)

Thanks again