Strange problem with integrals

[ManMan]

For some reason the result I get on some integrals like the one in the image is either double or half what it's supposed to be. For example on exercise 501 the result should be [MATH]2 sqrt(2) arctan ((x+3)/sqrt2)[/MATH] but I get double that.
I also added another one on top to show that I can do another one of these by applying the same method, the result on exercise 508 is correct. What am I doing wrong?

 

[skeeter]

[MATH]4\int \frac{\frac{1}{2}}{\frac{(x+3)^2}{2} + 1} \, dx[/MATH]
[MATH]\frac{4}{\sqrt{2}} \int \frac{\frac{1}{\sqrt{2}}}{\left(\frac{x+3}{\sqrt{2}}\right)^2 + 1} \, dx[/MATH]
[MATH]2\sqrt{2} \arctan\left(\frac{x+3}{\sqrt{2}}\right)+C[/MATH]
[MATH]2\sqrt{2} \arctan\left[\frac{\sqrt{2}(x+3)}{2}\right]+C[/MATH]

 

[Dr.Peterson]

At this step, you multiplied by 2:

Rather than multiplying by \(\sqrt{2}\) twice, you should multiply and divide.

 

[ManMan]

Now I get it, thanks

 

[lex]

You seem to skip a step at the end and that's where you make the mistake.
Going from the previous step to (*), you should ask, what did you multiply or divide the numerator by to get the desired value on top and then do 'the opposite' to the number outside the integral. You'll notice you got that bit wrong in both questions.
In 508 - you multiplied the top line by 2 (to change [MATH]\frac{1}{4}[/MATH] to [MATH]\frac{1}{2}[/MATH]), so you must divide outside by 2.
In 501 - you multiplied the top line by [MATH]\sqrt2[/MATH] (to change [MATH]\frac{1}{2}[/MATH] to [MATH]\frac{\sqrt2}{2}[/MATH]), so you must divide outside by [MATH]\sqrt2[/MATH].