- #1
Petrus
- 702
- 0
Hello MHB,
\(\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy\)
I start with subsitate \(\displaystyle u=x <=> du=dx\) and \(\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}\) so we got integrate by part that
\(\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx\)
and we got
\(\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1\)
Remember that we solve dx. Is this correct?
Regards,
\(\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy\)
I start with subsitate \(\displaystyle u=x <=> du=dx\) and \(\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}\) so we got integrate by part that
\(\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx\)
and we got
\(\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1\)
Remember that we solve dx. Is this correct?
Regards,