∫ 1 + x + x + x 2 x + 1 + x d x =

∫1+x+x+x2x+1+xdx=
  1. 121+x+c
  2. 23(1+x)32+c
  3. 1+x+c
  4. 2(1+x)32+c

Solution

∫1+x+x+x2x+1+xdx

=∫(1+x)2+x1+xx+1+xdx

=∫1+x[1+x+x]x+1+xdx

=∫(1+x)12dx

Using ∫xndx=xn+1n+1+c,

=(1+x)3232+c

=231+x32+c.

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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