The value of displaystyle ∫ (e√x/√x (1 + e2 √x))dx is equal to (where, C is the constant of integration)

Question

The value of displaystyle ∫ (e√x/√x (1 + e2 √x))dx is equal to (where, C is the constant of integration) (A) (tan)- 1 (2 e√x)+C (B) ln ((1 + e√x/1 - e√x))+C (C) 2(tan)- 1 (e√x)+C (D) ((tan)- 1 x)e√x+C

Tardigrade Admin · Accepted Answer

Let, I=∫ (e√x/√x(1+e2 √x)) d x Substituting e√x=t (e√x/2 √x) d x=d t I=2 ∫ (d t/1+t2) =2( tan -1 t)+C I=2 tan -1(e√x)+C

Q. The value of $\int \frac{e ^{x}}{x ( 1 + e ^{2 x} )} d x$ is equal to (where, $C$ is the constant of integration)

$(t an)^{- 1} (2 e^{x}) + C$

$l n (\frac{1 + e ^{x}}{1 - e ^{x}}) + C$

$2 (t an)^{- 1} (e^{x}) + C$

$((t an)^{- 1} x) e^{x} + C$

Let, $I = \int \frac{e ^{x}}{x ( 1 + e ^{2} x )} d x$
Substituting $e^{x} = t$ $\frac{e ^{x}}{2 x} d x = d t$
$I = 2 \int \frac{d t}{1 + t ^{2}}$
$= 2 (tan^{- 1} t) + C$
$I = 2 tan^{- 1} (e^{x}) + C$

Q. The value of ∫x​(1+e2x​)ex​​dx is equal to (where, C is the constant of integration)

(tan)−1(2ex​)+C

ln(1−ex​1+ex​​)+C

2(tan)−1(ex​)+C

((tan)−1x)ex​+C