If ∫x dxe2x−(x+1)2= cos–1 (f(x))+C, where Limx→0 f(x)=1 then
∫e2xf(x)dx= ex + C
∫e2xf(x)dx= xex + C
Limx→∞ ( 1+6f(x) ) ex3x= e2
Limx→0 f(x)−1x2=12
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(A) ∫x dxex1−(x+1ex)2
Put x+1ex=t⇒−xex dx=dt
∫−dt1−t2=cos−1(t)+C
⇒f(x)=x+1ex
⇒Limx→0 f(x)=1
Now, for check option
(B) ∫e2x(x+1ex) dx=x ex+C
(C) Limx→0 x+1−exex x2
=x+1−(1+x+x22!)ex · x2=−12
(D) Limx→∞ (1+6(x+1)ex)ex3x=eLimx→∞ ex3x(6(x+1)ex+1−1)=eLimx→∞ ex3x(6(x+1)ex)=e2