# Integrals on exam I couldn't answer

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

$$\int \frac{2}{(x-1)^2+1} dx$$

and

$$\int \frac{2}{\sqrt{1-(x-1)^2}} dx$$

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

Do you know how to do these integrals?

$$\int \frac{1}{x^2+1}\,dx$$

and

$$\int \frac{1}{\sqrt{1-x^2}}\,dx$$

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

$$\int \frac{1}{x^2+1}\,dx$$

and

$$\int \frac{1}{\sqrt{1-x^2}}\,dx$$
$$\int \frac{1}{\sqrt{1-x^2}}\,dx$$ = inverse sine

and

$$\int \frac{1}{x^2+1}\,dx$$ = inverse tangent

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

The factors of 2, you can just pull out of the integral, since they're constant multiplicative factors. As for the substitution, if you'd like to turn your integrals into the ones I gave, then you can see that you'd really like there to be x's where you have (x-1)'s. So that's what you have to substitute x = x-1. To be clear, introduce a new variable: u = x-1. Then du/dx = 1, so du = dx, and you can rewrite the integral purely in terms of u by substituting u for x-1 and du for dx.

$$\int \frac{2}{(x-1)^2+1} dx$$ 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

$$\int \frac{2}{\sqrt{1-(x-1)^2}} dx$$ 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

Yes, you're correct.

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

Thanks again