Help with integration

Iwasfrozen

could anyone help with this problem ?
[Integrate]tan^-1
This is what I've gotten so far:
let y=tan^-1
dy/dy= 1(1^2 + x^2)^-1 Log tables
y=[integrate]1(1^2 + x^2)^-1 = ln(1 + x^2)
So that leaves me with:
[Integrate]tan^-1 = [Integrate]ln(1 + x^2)
Thing is how do you integrate an ln ?
Thanks in advance.

I'm not sure what way you're doing it, but can't it be done by integration by parts?

Let u = arctan(x)
du = 1/(1 + x^2)

v = 1
dv = dx

Then,

[Integrate]udv = uv - [Integrate]vdu

It works out that way.

Thing is how do you integrate an ln ?

To integrate the natural log, you have to use integration by parts.

let u equal the log, and v equal 1.

Here: http://answers.yahoo.com/question/index?qid=1005121201875

LeixlipRed

Tan^-1 of what? It's meaningless without some variable or number.

Iwasfrozen

Tan^-1 of what? It's meaningless without some variable or number.

my bad, tan^-1x.
Thanks for the help. Jammy Doger.

Iwasfrozen

[Integrate]tan^-1x = [Integrate]tan-1x,1
[Integrate]udv = uv - [Integrate] vdu
[Integrate]tan^-1x,1 = (tan^-1)(x) - [Integrate](x)1/(1 + x^2)
=> xtan^-1 - (x)ln(1+x^2)
Is that right ? It seems abit easy for a part C.

It's all right except for the very last part.

To integrate x/(1+x^2) you need to use a substitution.

u = 1 + x^2
du = 2x dx
1/2 du = x dx

So the integral becomes 1/2[Integrate]du/u, which is (1/2)log_e(u), which is (1/2)log_e(1+x^2).

So your final integral becomes:

[Integrate] tan^-1(x) = x.tan^-1(x) - (1/2)log_e(1+x^2) + C.