integrate: (x4+1) /(x6+1)

∫x4+1x6+1dxOr we can write it as∫(x2+1)2-2x2(x2+1)(x4-x2+1)dx=∫x2+1(x4-x2+1)dx-2∫x2(x3)2+1dx=∫1/x2+1(x-1/x)2+1dx-2∫x2(x3)2+1dxSo let x-1/x = t, so (1-1/x2)dx =dt and x3=u, so 3x2dx =duSo it will change into=∫dt(t)2+1-23∫du(u)2+1=tan-1t -23tan-1u +cSo tan-1 (x-1/x) -23tan-1x3 +c

integrate: (x⁴+1) /(x⁶+1)

\int \frac{x^{4} + 1}{x^{6} + 1} d x O r w e c a n w r i t e i t a s \int \frac{(x^{2} + 1)^{2} - 2 x^{2}}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x = \int \frac{x^{2} + 1}{(x^{4} - x^{2} + 1)} d x - 2 \int \frac{x^{2}}{(x^{3})^{2} + 1} d x = \int \frac{1 / x^{2} + 1}{(x - 1 / x)^{2} + 1} d x - 2 \int \frac{x^{2}}{(x^{3})^{2} + 1} d x S o l e t x - 1 / x = t, s o (1 - 1 / x^{2}) d x = d t a n d x^{3} = u, s o 3 x^{2} d x = d u S o i t w i l l c h a n g e i n t o = \int \frac{d t}{(t)^{2} + 1} - \frac{2}{3} \int \frac{d u}{(u)^{2} + 1} = \tan^{- 1} t - \frac{2}{3} \tan^{- 1} u + c S o \tan^{- 1} (x - 1 / x) - \frac{2}{3} \tan^{- 1} x^{3} + c

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