Primitive function - smart substitution

[Rectifier]

The problem
$$ \int \frac{x}{\sqrt{x^2+2x+10}} \ dx $$

The attempt
$\int \frac{x}{\sqrt{x^2+2x+10}} \ dx = \int \frac{x}{\sqrt{(x+1)^2+9}} \ dx$

Is there any smart substitution I can make here to make this a bit easier to solve?

 

[Mark44]

The problem
$$ \int \frac{x}{\sqrt{x^2+2x+10}} \ dx $$

The attempt
$\int \frac{x}{\sqrt{x^2+2x+10}} \ dx = \int \frac{x}{\sqrt{(x+1)^2+9}} \ dx$

Is there any good substitution I can make here to make this a bit easier to solve?

Looks like this trig substitution might work, with $\tan(u) = \frac {x + 1} 3$

 

[Rectifier]

Oh, okay so I basically get $\int 3 \tan u - 1 \ du$ after that and that is much easier. Very elegant substitution indeed.

 

[fresh_42]

The list of formulas suggests (by looking at the result) to first do an integration by parts to get rid of the $x$ in the nominator, and then some logarithm with $z^2:=x^2+2x+10$. (But I only took a brief look.)

 

[Rectifier]

If we try to look for alternative substitutions;

That integral from my problem looks a lot like an standard-integral which I have in my book:

$\int \frac{1}{\sqrt{x^2+a}} = \ln | x + \sqrt{x^2+a} |$ but the only thing different is the x in the nominator so I guess I could do this by integrating by parts and removing the x as a derivative inside the integral but I will end up with a part with $...-\int {\ln | x + \sqrt{x^2+a}}| \ dx$ which is not much easier to solve

 

[Ray Vickson]

The problem
$$ \int \frac{x}{\sqrt{x^2+2x+10}} \ dx $$

The attempt
$\int \frac{x}{\sqrt{x^2+2x+10}} \ dx = \int \frac{x}{\sqrt{(x+1)^2+9}} \ dx$

Is there any smart substitution I can make here to make this a bit easier to solve?

Writing the numerator as $x = \frac{1}{2} (2x +2) -1$ we have
$$ \int \frac{x}{\sqrt{(x+1)^2+9}} \, dx = \frac{1}{2} \int \frac{d(x+1)^2}{\sqrt{(x+1)^2+9}} - \int \frac{dx}{\sqrt{(x+1)^2+9}}$$
The first one has the form $(1/2) \int du/\sqrt{u+9}$, while the second one has the form $\int du/\sqrt{u^2+9}$.