# How to use integration of parts on this?

How to use integration of parts on this?

$\int\frac{\sqrt{a+bu}}{u}du$

The answer should be summation of two parts in which one part still has integration sign on and one part is 2 $\sqrt {a+bu}$. In my textbook it was on the formula table but without any hint or solution

Second relevant question is how to integrate the first part that still had the integration sign on.

Don't ask me how to do start it by myself because i don't have money to hire a private tutor or to pay monthly fee to wolfram alpha...

• Respectfully, all I'm hearing is "don't ask me to do any work because I can't hire a tutor." Just because you can't hire a tutor doesn't mean you can't tell us what your thoughts are, what you tried, and what doesn't work. – Alfred Yerger Aug 21 '16 at 23:00
• Also, Wolfram Alpha is free on the web: wolframalpha.com – marty cohen Aug 22 '16 at 0:17
• @martycohen No complete step by step solution given without pay them... – Victor Aug 22 '16 at 1:31

## 2 Answers

Is necesary use integration of parts in this?. Make x= $\sqrt{a+bu}$, then dx=$\frac{1}{2u}du$, so: $$\int \frac{2x^{2}}{x^2-a} dx$$ $$= 2 \int 1+ \frac{a}{x^2-a}dx$$ $$= 2 (\int dx+ \int \frac{a}{x^2-a}dx)$$ $$= 2 (x+ a \int \frac{1}{x^2-a}dx)+k$$ k is the constant $$= 2 (x- a \int \frac{1}{\sqrt{a}^2-x^2}dx)+k$$ Im gonna use table, then: $$= 2 (x- a (\frac{1}{2\sqrt{a}}ln(\frac{\sqrt{a}+x}{\sqrt{a}-x}))+k$$ and remplace x: $$= 2 (\sqrt{a+bu}- a (\frac{1}{2\sqrt{a}}ln(\frac{\sqrt{a}+\sqrt{a+bu}}{\sqrt{a}-\sqrt{a+bu}}))+k$$

• wolframalpha.com/input/?i=%5Cint+sqrt%7Ba%2Bbu%7D%2Fu+du show a different answer... – Victor Aug 21 '16 at 22:55
• @Victor The inverse hyperbolic tangent can also be written in terms of a logarithm. – Ian Aug 21 '16 at 23:03
• @Victor, If you want to see if the primitive function is correct, you have to derive her. you have to get to the function you were integrating. – retro_var Aug 21 '16 at 23:08

Instead of integration by parts, you can use

$\displaystyle\int\frac{\sqrt{a+bu}}{u}\,du=\int\left(\frac{\sqrt{a+bu}}{u}-\frac{a}{u\sqrt{a+bu}}\right)du+\int\frac{a}{u\sqrt{a+bu}}\,du$

$\hspace{1.07 in}\displaystyle=\int\frac{(a+bu)-a}{u\sqrt{a+bu}}\,du+a\int\frac{1}{u\sqrt{a+bu}}\,du$

$\hspace{1.07 in}\displaystyle=\int\frac{b}{\sqrt{a+bu}}\,du+a\int\frac{1}{u\sqrt{a+bu}}\,du=2\sqrt{a+bu}+a\int\frac{1}{u\sqrt{a+bu}}\,du$.

(Probably the simplest way to find the last integral, as well as the original integral, is to use the substitution $x=\sqrt{a+bu}$ given by testpilot.)