∫(x^6 1x^2+1)dx here write x^6 1=x^4(x^2+1) (x^2+1)^2+2(x^2+1) 2 ∫(x^4(x^2+1) (x^2+1)^2+2(x^2+1) 2x^2+1)dx=∫ x^4dx ∫ (x^2+1)dx+2∫ dx 2∫dx1+x^2 ...

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Theorems on Integration

∫x6-1x2+1 dx

Question

$\int (\frac{x^{6} - 1}{x^{2} + 1}) d x$

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Solution

$\int (\frac{x^{6} - 1}{x^{2} + 1}) d x$
here write $x^{6} - 1 = x^{4} (x^{2} + 1) - (x^{2} + 1)^{2} + 2 (x^{2} + 1) - 2$
$\int (\frac{x^{4} (x^{2} + 1) - (x^{2} + 1)^{2} + 2 (x^{2} + 1) - 2}{x^{2} + 1}) d x = \int x^{4} d x - \int (x^{2} + 1) d x + 2 \int d x - 2 \int \frac{d x}{1 + x^{2}}$
$= \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {tan}^{- 1} x + C$

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