If displaystyle ∫ (x/x + 1 + ex) d x=px+qln |x + 1 + ex|+c, where c is the constant of integration, then p+q is equal to

Question

If displaystyle ∫ (x/x + 1 + ex) d x=px+qln |x + 1 + ex|+c, where c is the constant of integration, then p+q is equal to (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

displaystyle ∫ ((x + 1 + ex) - (ex + 1)/x + 1 + ex)dx= displaystyle ∫ (1 - ((ex + 1)/x + 1 + ex))dx=x-ln |x + 1 + ex|+c On comparing, p=1,q=-1 Hence, p+q=0

Q. If $\int \frac{x}{x + 1 + e ^{x}} d x = p x + ql n ∣ x + 1 + e^{x} ∣ + c,$ where $c$ is the constant of integration, then $p + q$ is equal to

$0$

$1$

$2$

$3$

$\int \frac{( x + 1 + e ^{x} ) - ( e ^{x} + 1 )}{x + 1 + e ^{x}} d x = \int (1 - \frac{( e ^{x} + 1 )}{x + 1 + e ^{x}}) d x = x - l n ∣ x + 1 + e^{x} ∣ + c$
On comparing,
$p = 1, q = - 1$
Hence, $p + q = 0$

Q. If ∫x+1+exx​dx=px+qln∣x+1+ex∣+c, where c is the constant of integration, then p+q is equal to

0

1

2

3

∫x+1+ex(x+1+ex)−(ex+1)​dx=∫(1−x+1+ex(ex+1)​)dx=x−ln∣x+1+ex∣+c On comparing, p=1,q=−1 Hence, p+q=0

Q. If $\int \frac{x}{x + 1 + e ^{x}} d x = p x + ql n ∣ x + 1 + e^{x} ∣ + c,$ where $c$ is the constant of integration, then $p + q$ is equal to

$0$

$1$

$2$

$3$

$\int \frac{( x + 1 + e ^{x} ) - ( e ^{x} + 1 )}{x + 1 + e ^{x}} d x = \int (1 - \frac{( e ^{x} + 1 )}{x + 1 + e ^{x}}) d x = x - l n ∣ x + 1 + e^{x} ∣ + c$
On comparing,
$p = 1, q = - 1$
Hence, $p + q = 0$