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

Question

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

$x l n (l n x) + c$

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

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

$e^{e^{x}} \cdot l n x + c$

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 $ln x = g (x)$
$∴ I = \int (f^{'} (x) g (x) + f (x) g^{'} (x)) d x$
$= f (x) \cdot g (x) + c$
$= e^{e^{x}} \cdot ln x + c$

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

xln(ln⁡x)+c

eex+c

xeex​+c

eex⋅lnx+c