całka podwójna

[ann_u]

Oblicz \(\displaystyle{ \int_{0}^{1} \int_{0}^{1} \frac{\ln(x^2+y^2)}{ \sqrt{1+x+y} }dxdy}\)

[Janusz Tracz]

Ukryta treść:    

\(\displaystyle{ \left(\frac{1}{36}+\frac{i}{36}\right) \left((-128+128 i)+(512-512 i) \sqrt{2}-(384-384 i) \sqrt{3}+24 \pi -12 (2-i)^{3/2} \log (2)+3 (2-i)^{3/2} \log (16)+24 (2-i)^{3/2} \log \left(\sqrt{3}+\frac{3}{\sqrt{2-i}}\right)-24 (2-i)^{3/2} \log \left(\sqrt{3}-\frac{3}{\sqrt{2-i}}\right)-24 (2-i)^{3/2} \log \left(\sqrt{2}+\frac{2}{\sqrt{2-i}}\right)+24 (2-i)^{3/2} \log \left(\sqrt{2}-\frac{2}{\sqrt{2-i}}\right)-(6+6 i) \log (8)-(3-6 i) \sqrt{2+i} \log (16)+(19-13 i) \log (64)+(24-24 i) \log \left(\sqrt{2}-1\right)-(24-24 i) \log \left(\sqrt{2}+1\right)+12 \log \left(2 \sqrt{2}+3\right)-(24-48 i) \sqrt{2+i} \log \left(\sqrt{3}-\frac{3}{\sqrt{2+i}}\right)-12 \log \left(1-\sqrt{2-i}\right)+12 \log \left(1+\sqrt{2-i}\right)-(24-48 i) \sqrt{2+i} \log \left(\sqrt{2}+\frac{2}{\sqrt{2+i}}\right)+12 i \log \left(\left(2 \sqrt{2}-3\right) \left(-1+\sqrt{2+i}\right)\right)-12 i \log \left(1+\sqrt{2+i}\right)+(36-36 i) \sqrt{3} \log (4)+(12-24 i) \sqrt{2+i} \log (2)+(24-48 i) \sqrt{2+i} \log \left(\sqrt{3}+\frac{3}{\sqrt{2+i}}\right)+(24-48 i) \sqrt{2+i} \log \left(\sqrt{2}-\frac{2}{\sqrt{2+i}}\right)-(96-96 i) \sinh ^{-1}(1)+(24-24 i) \tanh ^{-1}\left(\sqrt{2}\right)-(96-72 i) \sqrt{2+i} \tanh ^{-1}\left(\frac{1}{\sqrt{1+\frac{i}{2}}}\right)-(72-96 i) \sqrt{2-i} \tanh ^{-1}\left(\frac{1}{\sqrt{1-\frac{i}{2}}}\right)-(12-12 i) \sqrt{-22-4 i} \tanh ^{-1}\left(\sqrt{\frac{6}{5}+\frac{3 i}{5}}\right)+24 (2+i)^{3/2} \tanh ^{-1}\left(\sqrt{\frac{4}{5}-\frac{2 i}{5}}\right)-24 \tanh ^{-1}\left(\sqrt{2-i}\right)+24 i \tanh ^{-1}\left(\sqrt{2+i}\right)+(72-96 i) \sqrt{2-i} \tanh ^{-1}\left(\frac{1}{\sqrt{\frac{2}{3}-\frac{i}{3}}}\right)+(48-96 i) \sqrt{2+i} \tanh ^{-1}\left(\frac{1}{\sqrt{\frac{2}{3}+\frac{i}{3}}}\right)+(12-12 i) \sqrt{-22-4 i} \tanh ^{-1}\left(\sqrt{\frac{4}{5}+\frac{2 i}{5}}\right)\right)}\)