integrate : (x3+1)/(x4+1)dx

Let I=∫x3+1x4+1dxI=∫x3x4+1 dx +∫1x4+1 dxLet I1=∫x3x4+1 dx and I2=∫1x4+1 dxI1=∫x3x4+1 dxPut x4+1=t4x3 dx=dtx3 dx=dt4I1=14∫dtt=14 log t=14 log x4+1Now,I2=∫1x4+1 dx=∫1x2x2+1x2 dx=12∫2x2x2+1x2 dx=12∫1+1x2x2+1x2 dx-12∫1-1x2x2+1x2 dx=12∫1+1x2x-1x2+2 dx-12∫1-1x2x+1x2-2 dxPut x-1x=u in first integral and x+1x=v in second integral I2=12∫1u2+22 du -12∫1v2-22 dv=122 tan-1u2-142logv-2v+2+c=122 tan-1x2-12x-142 logx2-2x+1x2+2x+1 + cSo, I=14 log x4+1+122 tan-1x2-12x-142 logx2-2x+1x2+2x+1 + c ANS

integrate :
(x³+1)/(x⁴+1)dx

Let I = \int \frac{x^{3} + 1}{x^{4} + 1} dx I = \int \frac{x^{3}}{x^{4} + 1} dx + \int \frac{1}{x^{4} + 1} dx Let I_{1} = \int \frac{x^{3}}{x^{4} + 1} dx and I_{2} = \int \frac{1}{x^{4} + 1} dx I_{1} = \int \frac{x^{3}}{x^{4} + 1} dx Put x^{4} + 1 = t 4 x^{3} dx = dt x^{3} dx = \frac{dt}{4} I_{1} = \frac{1}{4} \int \frac{dt}{t} = \frac{1}{4} \log |t| = \frac{1}{4} \log |x^{4} + 1| Now, I_{2} = \int \frac{1}{x^{4} + 1} dx = \int \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx = \frac{1}{2} \int \frac{\frac{2}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 2} dx - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 2} dx Put x - \frac{1}{x} = u in first integral and x + \frac{1}{x} = v in second integral . I_{2} = \frac{1}{2} \int \frac{1}{u^{2} + {(\sqrt{2})}^{2}} du - \frac{1}{2} \int \frac{1}{v^{2} - {(\sqrt{2})}^{2}} dv = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} \log |\frac{v - \sqrt{2}}{v + \sqrt{2}}| + c = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c So, I = \frac{1}{4} \log |x^{4} + 1| + \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c ANS . . .

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