I'm trying to evaluate the following integral:

$$\int \frac{\sqrt{1+x}}{x}$$

It seems like that I need to use u substitution and partial fraction decomposition? Any tips/advice on how to solve this one? I can't figure it out.


3 Answers 3


Let $\sqrt{1+x} = u \implies 1+x = u^2 \implies dx = 2u du$. We then get $$I = \int \dfrac{\sqrt{1+x}}x dx = \int \dfrac{u}{u^2-1} 2u du = 2 \int \dfrac{u^2}{u^2-1} du = 2 \int \left(1 + \dfrac1{u^2-1}\right) du$$ Hence, $$I = 2u + \int \dfrac{du}{u-1} - \int \dfrac{du}{u+1} + c = 2 u + \ln \left(\left \vert \dfrac{u-1}{u+1} \right \vert\right) + \text{constant}$$

  • $\begingroup$ how does the fact $1+x\geq 0$ resolves the problem? we still don't know if $u$ is positive or not. $\endgroup$
    – palio
    Apr 22, 2013 at 15:08

Hint: Let $1+x=u^2$.${}{}{}{}{}{}{}$

You should end up needing $\displaystyle \int\frac{2u^2}{u^2-1}\,du$. Divide, getting $2+\frac{2}{u^2-1}$.


... just try putting $1+x = t^2$

  • $\begingroup$ m extremely sorry... its 1+x = t^2... sorry for typing mistake $\endgroup$
    – monalisa
    Mar 4, 2013 at 7:08
  • 6
    $\begingroup$ Dear Mona Lisa. MSE is like an on-line classroom. How would you like if, next time you ask your professor a question, he begins his answer by saying "it's very simple...how come you can't solve this?" Maybe keep that in mind. Thanks for your answer though. $\endgroup$ Mar 4, 2013 at 7:30
  • 5
    $\begingroup$ @gnometorule I heartily accept my fault. I had no intention of hurting anyone. i assure no such behaviour from my side in future. $\endgroup$
    – monalisa
    Mar 5, 2013 at 3:04

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