Integral is typical arctan type

[UrbanXrisis]

\int_{0}^{1}\frac{4}{1+x^2}

\int 4(1+x^2)^{-1}

u=1+x^2

du=2x

\frac{2du}{x}=4dx

2\int \frac{u^{-1}}{x} dx

2ln(ux)

=2ln(x+x^3)|_{0}^{1}

=2ln(1+1^3)-2ln(0+0^3)

I know I did something wrong, any ideas?

 

[dextercioby]

Yes,your integral is typical \arctan type...You should know this

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

by heart...

Daniel.

 

[Data]

If you make a substitution, you need to arrange it such that the resulting expression does not contain the variable you substituted for.

Here, you had

\int 4(1+x^2)^{-1} \ dx,

and made the substitution

u = 1 + x^2 \Longrightarrow du = 2x \ dx \Longrightarrow \frac{du}{2x} = dx

you now need to express dx in terms of only u, ie. with no x's on the left side. You can do this by noting

2x = 2\sqrt{(x^2 + 1) - 1} = 2\sqrt{u - 1}

so you get

dx = \frac{du}{2\sqrt{u - 1}}

so your integral is

\int \frac{4}{2u\sqrt{u-1}} \ du = \int \frac{2}{u\sqrt{u-1}} \ du

which is not very nice. I would advise seeing if you can find a better strategy to start with.

 

[SpaceTiger]

\frac{2du}{x}=4dx

2\int \frac{u^{-1}}{x} dx

When you make a substitution, you want to express everything in terms of the new variable before integrating. Try instead the substitution:

x=tan(u)

 

[UrbanXrisis]

I've never really learned about acrtan... how does tan(u) fit into this equation?

 

[dextercioby]

If u know the substitution metod and apply it for the given function (tangent) u'll learn.

Daniel.

 

[Data]

Subsituting x = \tan{u} \Longrightarrow dx = \sec^2 u \ du simplifies the integral completely. Substituting these in gives

\int \frac{4}{1+x^2} \ dx = \int \frac{4}{1 + \tan^2 u} \sec^2 u \ du

Then all you need to know is

\sec^2 u - \tan^2 u = 1

which you can prove on your own, and I'm sure you can finish from there

 

[dextercioby]

Whozum,you're wrong.

Daniel.

EDIT:Yoiu saw it & deleted the post...

 

[Jameson]

This is a fairly straightfoward arctangent rule as others have said.

\int \frac{4}{1 + x^2}dx = 4 \int\frac{1}{1 + x^2}dx = 4\arctan{x} + C

The rule for tangent inverses is \int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C

 

[Data]

don't forget your constants~

 

[Jameson]

Well, since this was a definite integral I thought I'd let the poster apply the bounds, but yes, you are correct. I'll change them...