Question

Va rog ajutor la o integrala. Multumesc.

Answer 1

[tex]I =\displaystyle \int \dfrac{1}{x\sqrt{1+x}} \, dx \\ \\ \sqrt{1+x} = t \Rightarrow 1+x = t^2 \Rightarrow x = t^2-1 \Rightarrow dx = 2t \, dt \\ x = t^2 -1 \\ \\ I = \int\dfrac{1}{(t^2-1)\cdot t} \cdot 2t \, dt = 2 \int \dfrac{1}{t^2-1^2} \, dt = 2\cdot \dfrac{1}{2\cdot 1} \ln \left|\dfrac{1+t}{1-t}\right| +C = \\ \\ = \ln \left|\dfrac{1-t}{1+t}\right|+C = \ln\left|\dfrac{1-\sqrt{1+x}}{1+\sqrt{1+x}}\right| +C\\ \\ \\\Rightarrow \int \dfrac{1}{x\sqrt{1+x}} \, dx = \ln\left|\dfrac{1-\sqrt{1+x}}{1+\sqrt{1+x}}\right| +C[/tex]


$\displaystyle \text{M-am folosit de formula:} \\ \\ \int \dfrac{1}{u^2-a^2} \, du = \dfrac{1}{2a} \ln \left|\dfrac{1+u}{1-u}\right| +C$

Answer 2

$I= \int\ {1/x \sqrt{x+1} } \, dx$
Facem schimbarea de variabila:
$\sqrt{x+1}=t$
Ridicand la patrat obtinem:
$x+1=t^{2}$
Deci vom avea: dx=2tdt
Daca $x+1=t^{2} =\ \textgreater \ x=t^{2}-1$.
Si integrala I devine:
$I=\int\ {2t/(t^2-1)*t} \, dt=2\int\ {t/(t^2-1)*t} \, dt= se ~simplifica~t~cu~t=$$2 \int\ {1/t^{2}-1} \, dt =2*1/2*ln|(1-t)/(1+t)|+C=$$ln|(1-t)/(1+t)|+C$.
Revenind la notatia initiala avem:
$I=ln|(1- \sqrt{x+1} )/(1+ \sqrt{x+1} )|+C$