Engineering
Mathematics
Integration by Substitution
Question

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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Solution

(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