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.