Factoring denominator of an integral

[dlthompson81]

Homework Statement

Ok, this is a pretty simple integral, but I'm having trouble with the factoring.

[itex]\int \frac{1}{x^{2}+2}[/itex]

According to the book, the answer is:

[itex]\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2})}[/itex]

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

[itex]\int\frac{1}{x^{2}+1}[/itex]

I did this by factoring out a [itex]\sqrt{2}[/itex]:

[itex]\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2})}[/itex]

But when you convert the [itex]\frac{x^{2}}{\sqrt{2}}[/itex] to [itex](\frac{x}{\sqrt{2}})^{2}[/itex] the [itex]\sqrt{2}[/itex] on the outside of the factor doesn't cancel out the one being squared. I'm kind of lost here.

Squaring the bottom term produces a 2 which doesn't cancel with the [itex]\sqrt{2}[/itex] on the outside of the parenthesis, and changing the term to [itex]4\sqrt{2}[/itex] which would square and cancel isn't in the given answer.

 

[Mark44]

Homework Statement

Ok, this is a pretty simple integral, but I'm having trouble with the factoring.

[itex]\int \frac{1}{x^{2}+2}[/itex]

Don't forget dx! You are consistently omitting it in your integrals. It is crucial in problems like this.

The simplest way to do this is to use a trig substitution. Draw a right triangle with an acute angle θ. Label the altitude as x and the base as 2. From this we see that tan(θ) = x/2, so 2sec²(θ)dθ = dx.

Replace all expressions with x and dx in your original integral, and you'll have an easier one to integrate.

According to the book, the answer is:

[itex]\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2}} )[/itex]

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

[itex]\int\frac{1}{x^{2}+1}[/itex]

I did this by factoring out a [itex]\sqrt{2}[/itex]:

[itex]\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2}})[/itex]

 

[Mark44]

I should mention that you can factor the denominator and use the method of partial fraction decomposition. x² + 2 factors into (x + i√2)(x - i√2).

 

[HallsofIvy]

Homework Statement

Ok, this is a pretty simple integral, but I'm having trouble with the factoring.

[itex]\int \frac{1}{x^{2}+2}[/itex]

According to the book, the answer is:

[itex]\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2})}[/itex]

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

[itex]\int\frac{1}{x^{2}+1}[/itex]

I did this by factoring out a [itex]\sqrt{2}[/itex]:

[itex]\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2})}[/itex]

If you need [itex]x^2+ [/itex] rather than [itex]x^2+ 2[/itex] you surely don't want [itex]x^2+ \sqrt{2}[/itex]! Don't factor out [itex]\sqrt{2}[/itex], factor out [itex]2[/itex]:
[tex]\frac{1}{2}\int \frac{dx}{\frac{x^2}{2}+ 1}[/tex]
Now let [itex]u= x/\sqrt{2}[/itex]. As Mark44 says,don't forget the dx! [itex]x= \sqrt{2}u[/itex] so [itex]dx= \sqrt{2}du[/itex]

invert the [itex]\frac{x^{2}}{\sqrt{2}}[/itex] to [itex](\frac{x}{\sqrt{2}})^{2}[/itex] the [itex]\sqrt{2}[/itex] on the outside of the factor doesn't cancel out the one being squared. I'm kind of lost here.

Squaring the bottom term produces a 2 which doesn't cancel with the [itex]\sqrt{2}[/itex] on the outside of the parenthesis, and changing the term to [itex]4\sqrt{2}[/itex] which would square and cancel isn't in the given answer.