What are the necessary steps to do this substitution?

Question

I can't handle this excersise:

$\int\sqrt{1+\frac{1}{3x}}\frac{dx}{x^2}$

And the answer is:

$2(1+\frac{1}{3x})\sqrt{1+\frac{1}{3x}}+c$

Answer 1

Use the substitution $u = 1+\frac{1}{3x}$ and $du = -\frac{1}{3x^2}dx$ or $-3du=\frac{1}{x^2}dx$.

Your integrad becomes $$\sqrt{u}\cdot (-3)du$$

This can be integrated by using the power law for integration as $\sqrt{u}=u^{\frac{1}{2}}$.

Answer 2

Try $$ u = 1+\frac{1}{3x} $$ but this does not lead to the answer you get $$ -2\left(1+\frac{1}{3x}\right)\sqrt{1+\frac{1}{3x}} + C $$

Answer 3

The gist is to "recognize" any involved derivatives and compositions and use whenever opportune the fundamental theorem of calculus and change of variables theorem. To be familiar with these requires experience, of course.

In this case, we have $$ \int_{x} \sqrt{1+\frac{1}{3x}} \frac{1}{x^{2}} = \int_{x} \sqrt{1 + \frac{1}{3x}}D(1 + \frac{1}{3x})(-3) = -3\int_{u:= 1 + \frac{1}{3x}} \sqrt{u} = -2\int_{u} Du^{3/2} = -2u^{3/2} + C = -2(1+ \frac{1}{3x})^{3/2} + C. $$