$int_{0}^{frac{5}{4}} frac{x+sqrt{x+1}}{2x-3sqrt{x+1}} , dx$












3












$begingroup$


I have to solve this integral:



$$int_{0}^{frac{5}{4}} frac{x+sqrt{x+1}}{2x-3sqrt{x+1}} , dx$$
I tried with the substitution $u=sqrt{x+1}=g^{-1}(x)Rightarrow x=u^2-1=g(u) Rightarrow dx= g'(u)du=2udu$



then the integral becomes:
$$int frac{2u^3+2u^2-2u}{2u^2-3u-2} , du$$



Dividing the numerator :
$$int (u+frac{5}{2} ), du+int frac{frac{15}{2}u+5}{2u^2-3u-2} , du=$$
$$int (u+frac{5}{2} ), du+frac{5}{4}int frac{3u+2}{(u-2)(u+frac{1}{2})} , du$$
$$=int (u+frac{5}{2} ), du+frac{5}{4}int frac {16}{5}frac{1}{(u-2)} , du+frac{5}{4}int (-frac {1}{5})frac{1}{(u+frac{1}{2})} , du=$$
$$frac{u^2}{2}+frac{5}{2}u+4log|u-2|-frac{1}{4}log|u+frac{1}{2}|=$$
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log|sqrt{x+1}-2|-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|+C$$



The right result for the indefinite integral is:
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log(2-sqrt{x+1})-frac{1}{4}log(2sqrt{x+1}+1)+C$$



and I don't understand why it is writtem $(2-sqrt{x+1})$ instead of $|sqrt{x+1}-2|$ and the difference in the last integral $-frac{1}{4}log(2sqrt{x+1}+1)$ and $-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|$



Can someone explain me why?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
    $endgroup$
    – Zubin Mukerjee
    Jan 17 at 23:12










  • $begingroup$
    @ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
    $endgroup$
    – Anne
    Jan 17 at 23:27












  • $begingroup$
    @Fabio yes. indeed, I've edited the text
    $endgroup$
    – Anne
    Jan 17 at 23:29
















3












$begingroup$


I have to solve this integral:



$$int_{0}^{frac{5}{4}} frac{x+sqrt{x+1}}{2x-3sqrt{x+1}} , dx$$
I tried with the substitution $u=sqrt{x+1}=g^{-1}(x)Rightarrow x=u^2-1=g(u) Rightarrow dx= g'(u)du=2udu$



then the integral becomes:
$$int frac{2u^3+2u^2-2u}{2u^2-3u-2} , du$$



Dividing the numerator :
$$int (u+frac{5}{2} ), du+int frac{frac{15}{2}u+5}{2u^2-3u-2} , du=$$
$$int (u+frac{5}{2} ), du+frac{5}{4}int frac{3u+2}{(u-2)(u+frac{1}{2})} , du$$
$$=int (u+frac{5}{2} ), du+frac{5}{4}int frac {16}{5}frac{1}{(u-2)} , du+frac{5}{4}int (-frac {1}{5})frac{1}{(u+frac{1}{2})} , du=$$
$$frac{u^2}{2}+frac{5}{2}u+4log|u-2|-frac{1}{4}log|u+frac{1}{2}|=$$
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log|sqrt{x+1}-2|-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|+C$$



The right result for the indefinite integral is:
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log(2-sqrt{x+1})-frac{1}{4}log(2sqrt{x+1}+1)+C$$



and I don't understand why it is writtem $(2-sqrt{x+1})$ instead of $|sqrt{x+1}-2|$ and the difference in the last integral $-frac{1}{4}log(2sqrt{x+1}+1)$ and $-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|$



Can someone explain me why?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
    $endgroup$
    – Zubin Mukerjee
    Jan 17 at 23:12










  • $begingroup$
    @ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
    $endgroup$
    – Anne
    Jan 17 at 23:27












  • $begingroup$
    @Fabio yes. indeed, I've edited the text
    $endgroup$
    – Anne
    Jan 17 at 23:29














3












3








3





$begingroup$


I have to solve this integral:



$$int_{0}^{frac{5}{4}} frac{x+sqrt{x+1}}{2x-3sqrt{x+1}} , dx$$
I tried with the substitution $u=sqrt{x+1}=g^{-1}(x)Rightarrow x=u^2-1=g(u) Rightarrow dx= g'(u)du=2udu$



then the integral becomes:
$$int frac{2u^3+2u^2-2u}{2u^2-3u-2} , du$$



Dividing the numerator :
$$int (u+frac{5}{2} ), du+int frac{frac{15}{2}u+5}{2u^2-3u-2} , du=$$
$$int (u+frac{5}{2} ), du+frac{5}{4}int frac{3u+2}{(u-2)(u+frac{1}{2})} , du$$
$$=int (u+frac{5}{2} ), du+frac{5}{4}int frac {16}{5}frac{1}{(u-2)} , du+frac{5}{4}int (-frac {1}{5})frac{1}{(u+frac{1}{2})} , du=$$
$$frac{u^2}{2}+frac{5}{2}u+4log|u-2|-frac{1}{4}log|u+frac{1}{2}|=$$
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log|sqrt{x+1}-2|-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|+C$$



The right result for the indefinite integral is:
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log(2-sqrt{x+1})-frac{1}{4}log(2sqrt{x+1}+1)+C$$



and I don't understand why it is writtem $(2-sqrt{x+1})$ instead of $|sqrt{x+1}-2|$ and the difference in the last integral $-frac{1}{4}log(2sqrt{x+1}+1)$ and $-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|$



Can someone explain me why?






real-analysis







share|cite|improve this question











$endgroup$




I have to solve this integral:



$$int_{0}^{frac{5}{4}} frac{x+sqrt{x+1}}{2x-3sqrt{x+1}} , dx$$
I tried with the substitution $u=sqrt{x+1}=g^{-1}(x)Rightarrow x=u^2-1=g(u) Rightarrow dx= g'(u)du=2udu$



then the integral becomes:
$$int frac{2u^3+2u^2-2u}{2u^2-3u-2} , du$$



Dividing the numerator :
$$int (u+frac{5}{2} ), du+int frac{frac{15}{2}u+5}{2u^2-3u-2} , du=$$
$$int (u+frac{5}{2} ), du+frac{5}{4}int frac{3u+2}{(u-2)(u+frac{1}{2})} , du$$
$$=int (u+frac{5}{2} ), du+frac{5}{4}int frac {16}{5}frac{1}{(u-2)} , du+frac{5}{4}int (-frac {1}{5})frac{1}{(u+frac{1}{2})} , du=$$
$$frac{u^2}{2}+frac{5}{2}u+4log|u-2|-frac{1}{4}log|u+frac{1}{2}|=$$
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log|sqrt{x+1}-2|-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|+C$$



The right result for the indefinite integral is:
$$frac{x+1}{2}+frac{5}{2} sqrt{x+1}+4log(2-sqrt{x+1})-frac{1}{4}log(2sqrt{x+1}+1)+C$$



and I don't understand why it is writtem $(2-sqrt{x+1})$ instead of $|sqrt{x+1}-2|$ and the difference in the last integral $-frac{1}{4}log(2sqrt{x+1}+1)$ and $-frac{1}{4}log|sqrt{x+1}-frac{1}{2}|$



Can someone explain me why?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 23:33







Anne

















asked Jan 17 at 23:03









AnneAnne

790419




790419












  • $begingroup$
    What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
    $endgroup$
    – Zubin Mukerjee
    Jan 17 at 23:12










  • $begingroup$
    @ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
    $endgroup$
    – Anne
    Jan 17 at 23:27












  • $begingroup$
    @Fabio yes. indeed, I've edited the text
    $endgroup$
    – Anne
    Jan 17 at 23:29


















  • $begingroup$
    What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
    $endgroup$
    – Zubin Mukerjee
    Jan 17 at 23:12










  • $begingroup$
    @ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
    $endgroup$
    – Anne
    Jan 17 at 23:27












  • $begingroup$
    @Fabio yes. indeed, I've edited the text
    $endgroup$
    – Anne
    Jan 17 at 23:29
















$begingroup$
What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
$endgroup$
– Zubin Mukerjee
Jan 17 at 23:12




$begingroup$
What is the value of $sqrt{x+1}$ when $x$ is between $0$ and $5/4$? Is it less than or greater than $2$?
$endgroup$
– Zubin Mukerjee
Jan 17 at 23:12












$begingroup$
@ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
$endgroup$
– Anne
Jan 17 at 23:27






$begingroup$
@ZubinMnhaveukerjee I think I have understood: when $0<x< frac{5}{4} $, $sqrt{x+1}<2$ and then $|sqrt{x+1}-2|=2-sqrt{x+1}$
$endgroup$
– Anne
Jan 17 at 23:27














$begingroup$
@Fabio yes. indeed, I've edited the text
$endgroup$
– Anne
Jan 17 at 23:29




$begingroup$
@Fabio yes. indeed, I've edited the text
$endgroup$
– Anne
Jan 17 at 23:29










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