ngoclam_bg

dat t=$\sqrt{x}+\sqrt{x+1}$

suy ra x=$\dfrac{1}{4}\left ( t-\dfrac{1}{t} \right )^{2}$

I=$\dfrac{1}{2}\int \dfrac{t^{4}-1}{t^{4}+t^{3}}dt$

tinh toan cho ta I=$\dfrac{t}{2}-\dfrac{1}{2}\ln t-\dfrac{1}{2t}+\dfrac{1}{4t^{2}}$=



$\dfrac{\sqrt{x}+\sqrt{1+x}}{2}-\dfrac{1}{2}\ln \left |

\sqrt{x}+\sqrt{1+x} \right |-\dfrac{1}{2\left (

\sqrt{x}+\sqrt{1+x}\right )}+\dfrac{1}{4\left (

\sqrt{x}+\sqrt{1+x}})^{2}$