Integral of (x^3)/(x^2 + 1)

[@d-hat-vr2002]

If you didn't spot the trick of factoring the numerator, a first pass using polynomial long division would also be a good first step for this integral. At 5:59 - the absolute value bars aren't needed in the ln, since x²+1 is always positive.

[@ناصريناصر-س4ب]

(x³)/(x²+1)=(x³+x-x)/(x²+1)=x-(x)/(x²+1)=x-(1/2)*[(2x)/(x²+1)] So the integral of the function is equal to x²/2-(1/2)ln(x²+1)+c

[@mathpro926]

Nice integral

[@gelbkehlchen]

∫x³/(x²+1)*dx = ------------------ Substitution: u = x²+1 x² = u-1 du = 2x*dx dx = du/(2x) ------------------ = 1/2*∫x³/u*du/x = 1/2*∫x²/u*du = 1/2*∫(u-1)/u*du = 1/2*[∫u/u*du-∫1/u*du] = 1/2*[∫du-∫1/u*du] = 1/2*[u-ln|u|+K] = 1/2*[x²+1-ln|x²+1|]+C