The indefinite integral ∫ eex (x ex ⋅ ln x+1/x) d x simplifies to (where, c is the constant of integration)

Question

The indefinite integral ∫ eex (x ex ⋅ ln x+1/x) d x simplifies to (where, c is the constant of integration) (A) x ln ( ln x)+c (B) eex+c (C) (eex/x)+C (D) eex ⋅ ln x+c

Tardigrade Admin · Accepted Answer

Given integral is I=∫ eex ⋅ ex ⋅ ln x+(eex/x) d x Let, eex=f x and ln x=g x ∴ I=∫ f prime x g x+f x g prime x d x =f x ⋅ g x+c =eex ⋅ ln x+c

Q. The indefinite integral $\int e^{e^{x} \frac{x e ^{x} \cdot l n x + 1}{x}} d x$ simplifies to (where, $c$ is the constant of integration)

$x ln (ln x) + c$

$e^{e^{x}} + c$

$\frac{e ^{e^{x}}}{x} + C$

$e^{e^{x}} \cdot ln x + c$

Given integral is $I = \int e^{e^{x}} \cdot e^{x} \cdot ln x + \frac{e ^{e^{x}}}{x} d x$
Let, $e^{e^{x}} = f x$ and $ln x = gx$
$∴ I = \int f^{'} xgx + f x g^{'} x d x$
$= f x \cdot gx + c$
$= e^{e^{x}} \cdot ln x + c$

Q. The indefinite integral ∫eexxxex⋅lnx+1​dx simplifies to (where, c is the constant of integration)

xln(lnx)+c

eex+c

xeex​+C

eex⋅lnx+c