∫2x3x2+1dx
u=x2
du=2xdx
∫uu+1du=∫u+1−1u+1du=∫1−1u+1du
how should I continue? is there an algotherm for integrating rational functions?
Answer
Notice, here is another simple method, ∫2x3x2+1 dx
∫2x3+2x−2xx2+1 dx
=∫2x(x2+1)−2xx2+1 dx
=∫2x dx−∫2xx2+1 dx
=2∫x dx−∫d(x2+1)x2+1
=x2−ln(x2+1)+C
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