Evaluate∫ex2x-2x2dx=
exx+c
ex2x2+c
2exx+c
2exx2+c
Explanation for the correct option:
Finding the integral:
∫ex2x-2x2dx=2∫ex1x-1x2dx......Taking2common=2∫ex1xdx-2∫exx2dxByusingintegrationbyparts=2·1x·∫ex-∫d1xdx∫exdx-2∫exx2dx+c=2·1x·ex-∫-1x2exdx-2∫exx2dx+c=2·1x·ex+2∫exx2dx-2∫exx2dx+c=2exx+c ∫u.vdx=u∫vdx-∫(u'∫vdx)
Hence, option (C) is correct.