displaystyle ∫ (ln (x + 1) - ln â¡ x/x (x + 1))dx is equal to (where C is an arbitarary constant)

Question

displaystyle ∫ (ln (x + 1) - ln â¡ x/x (x + 1))dx is equal to (where C is an arbitarary constant) (A) -(1/2)([ln ((x + 1/x))])2+C (B) C - [( ln (x + 1) )2 - (ln ⁡ x)2] (C) -ln [ln â¡ ((x + 1/x))]+C (D) -ln ((x + 1/x))+C

Tardigrade Admin · Accepted Answer

Put ln (x + 1)-ln â¡ x=t ⇒ (1/x + 1)-(1/x)=(d t/d x)⇒ (x - (x + 1)/x (x + 1))=(d t/d x) ⇒ (- d x/x (x + 1))=dt⇒ (d x/x (x + 1))=- dt so question becomes - displaystyle ∫ ât dt=-(t2/2)+C

Q. $\int \frac{l n ( x + 1 ) - l n x}{x ( x + 1 )} d x$ is equal to (where $C$ is an arbitarary constant)

$- \frac{1}{2} ([l n (\frac{x + 1}{x})])^{2} + C$

$C - [({l n (x + 1)})^{2} - (l n x)^{2}]$

$- l n [l n (\frac{x + 1}{x})] + C$

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

Q. ∫x(x+1)ln(x+1)−ln⁡x​dx is equal to (where C is an arbitarary constant)

−21​([ln(xx+1​)])2+C

C−[({ln(x+1)})2−(ln⁡x)2]

−ln[ln⁡(xx+1​)]+C

−ln(xx+1​)+C

Put ln(x+1)−ln⁡x=t ⇒x+11​−x1​=dxdt​⇒x(x+1)x−(x+1)​=dxdt​ ⇒x(x+1)−dx​=dt⇒x(x+1)dx​=−dt so question becomes −∫‍tdt=−2t2​+C

Q. $\int \frac{l n ( x + 1 ) - l n x}{x ( x + 1 )} d x$ is equal to (where $C$ is an arbitarary constant)

$- \frac{1}{2} ([l n (\frac{x + 1}{x})])^{2} + C$

$C - [({l n (x + 1)})^{2} - (l n x)^{2}]$

$- l n [l n (\frac{x + 1}{x})] + C$

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