Indefinate Integral 1/(x^2 +1)

[Calu]

Homework Statement

This is the very last piece of an integration using partial fractions question. I have to integrate 1/x²+1.

Homework Equations

Tan²x + 1 = Sec²x

The Attempt at a Solution

Substitute tanu=x:

1/tan²u + 1 = 1/sec²u = cos²u

which leaves me with int [ cos²u ] .dx I'm not sure what to do from here. Or if what I've done is correct.

 

[gopher_p]

Homework Statement

This is the very last piece of an integration using partial fractions question. I have to integrate 1/x²+1.

Homework Equations

Tan²x + 1 = Sec²x

The Attempt at a Solution

Substitute tanu=x:

1/tan²u + 1 = 1/sec²u = cos²u

which leaves me with int [ cos²u ] .dx I'm not sure what to do from here. Or if what I've done is correct.

Do you know a function whose derivative is $\frac{1}{x^2+1}$?

 

[D H]

which leaves me with int [ cos²u ] .dx

You haven't complete the u-substitution. You need to express dx as (something)*du. You used the u-substitution tan(u)=x, so what does that mean dx is?

 

[Calu]

You haven't complete the u-substitution. You need to express dx as (something)*du. You used the u-substitution tan(u)=x, so what does that mean dx is?

I'm not sure, that's the bit I'm having difficulty with.

 

[D H]

What is $\frac {d}{du} (\tan u)$ ?

 

[Calu]

I'm not sure, that's the bit I'm having difficulty with.

I think I have it. I have

d/dx(tanu)=d/dx(x)
du/dx(sec²u)=1
du/dx=1/sec²u
dx=sec²u .du

but I believe this to be incorrect, as that leaves me with

∫cos²x.sec²x.du = ∫1 .du = u.

 

[Calu]

What is $\frac {d}{du} (\tan u)$ ?

I believe that is sec²u, but I don't see how it relates to dx.

 

[LCKurtz]

I think I have it. I have

d/dx(tanu)=d/dx(x)
du/dx(sec²u)=1
du/dx=1/sec²u
dx=sec²u .du

but I believe this to be incorrect, as that leaves me with

∫cos²x.sec²x.du = ∫1 .du = u.

Aside from the missing constant of integration, that is correct. You started with $x=\tan u$. What do you get if you solve for $u$ to get your answer in terms of $x$?

 

[Calu]

Aside from the missing constant of integration, that is correct. You started with $x=\tan u$. What do you get if you solve for $u$ to get your answer in terms of $x$?

Ahh, I see now. arctanx=u. So ∫1/x² + 1 .dx = arctanx +c. Thanks.

 

[Mark44]

You need to use parentheses when there are multiple terms in the top or bottom of a fraction.

I have to integrate 1/x²+1.

Write this as 1/(x² + 1). What you wrote is (1/x²) + 1, which isn't what you meant.

1/tan²u + 1 = 1/sec²u = cos²u

Ahh, I see now. arctanx=u. So ∫1/x² + 1 .dx = arctanx +c. Thanks.

Both examples above should be 1/(tan²(u) + 1).