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So I want to integrate $\frac{1}{x^2+10x+50}$ and when I calculate it with an online calculator, I get $\frac{1}{x^2+10x+50}=\frac{1}{(x+5)^2+25} => u=\frac{x+5}{5}=>\frac{du}{dx}=\frac{1}{5}$

so that $\frac{1}{x^2+10x+50}=\frac{1}{5}\frac{1}{u^2+1}$

As far as I understood, the most common substitution I see is that the term is transformed into a function times the derivative of said function, which is when you can substitute the function and the derivative disappears. Here I can't wrap my head around as to what happens mathematically in this substitution - any help?

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    How does it become $(x^2+5)+25$? Do you mean $(x+5)^2+25$? – 2017-02-28
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    A typo that I'll fix – 2017-02-28

2 Answers 2

1

The arc tangent integral which often is to learn by heart is actually a short-cut for the arcus tangens expanded to complex numbers which can be defined as:

$$\arctan(x) = \frac{i}{2}\left(\phantom{\frac{}{}}\ln(1-ix) - \ln(1+ix)\phantom{\frac{}{}}\right)$$

In combination with the traditional fraction-decomposition which you would do whenever you encounter a fraction between two polynomials $\frac{P(x)}{Q(x)}$ to try and reduce to lowest degrees:

$$\frac{1}{1+x^2} = \left(\phantom{\frac{1}{1}}\text{conjugate rule}\phantom{\frac{1}{1}}\right) = \frac{1}{(1+ix)(1-ix)} = \left(\phantom{\frac{1}{1}} \text{partial fraction}\phantom{\frac{1}{1}}\right) = \frac{1}{1+ix}-\frac{1}{1-ix}$$

Then the denominators of that expression will have the natural logs as their integrals.

You probably won't encounter this until your first course (analysis) in one complex variable. But it's good to know it's not any particular "special magic" about the arc tan function.

2

You placed your square in the wrong places. You should instead have

$$\frac1{x^2+10x+50}=\frac1{(x+5)^2+25}$$

Then let $u=x+5$, and it simplifies to

$$\frac1{u^2+5^2}$$

Which is a simple arctangent integral:

$$\int\frac1{u^2+5^2}\ du=\frac15\arctan\left(\frac u5\right)+c=\frac15\arctan\left(\frac{x+5}5\right)+c$$

where in general we have

$$\int\frac1{u^2+a^2}\ du=\frac1{|a|}\arctan\left(\frac u{|a|}\right)+c$$

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    There is supposed to be an $\frac{1}{5}$ multiplier to the function according to calculators, though - where could that come from? – 2017-02-28
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    It is not needed. Take the derivative of the last line to prove it without ever performing that substitution. – 2017-02-28
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    Does that mean the same thing as [this](https://www.wolframalpha.com/input/?i=integral+of+1%2F(x%5E2%2Ba%5E2)) or something else? This is what is bugging me a bit, basically. – 2017-03-01
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    Oh, that is my bad. I always forget which inverse trig functions have a factor out front and which doesn't. – 2017-03-01