Factoring denominator of an integral

[dlthompson81]

Homework Statement

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

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

According to the book, the answer is:

\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2})}

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

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

I did this by factoring out a \sqrt{2}:

\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2})}

But when you convert the \frac{x^{2}}{\sqrt{2}} to (\frac{x}{\sqrt{2}})^{2} the \sqrt{2} 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 \sqrt{2} on the outside of the parenthesis, and changing the term to 4\sqrt{2} 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.

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

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[/color] in your original integral, and you'll have an easier one to integrate.

According to the book, the answer is:

\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2}} )

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

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

I did this by factoring out a \sqrt{2}:

\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2}})

 

[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.

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

According to the book, the answer is:

\frac{1}{\sqrt{2}} tan^{-1}(\frac{x}{\sqrt{2})}

Homework Equations

The Attempt at a Solution

So I need to get it in the form of:

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

I did this by factoring out a \sqrt{2}:

\frac{1}{\sqrt{2}(\frac{x^{2}}{\sqrt{2}}+\sqrt{2})}

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

invert the \frac{x^{2}}{\sqrt{2}} to (\frac{x}{\sqrt{2}})^{2} the \sqrt{2} 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 \sqrt{2} on the outside of the parenthesis, and changing the term to 4\sqrt{2} which would square and cancel isn't in the given answer.