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int (1)/(e^x+e^-x)^2 dx...

`int (1)/(e^x+e^-x)^2 dx`

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int(1)/(e^(x)+e^(-x))dx

int(1)/(3e^(x)+2e^(-x))dx=

Column I, a) int(e^(2x)-1)/(e^(2x)+1)dx is equal to b) int1/((e^x+e^(-x))^2)dx is equal to c) int(e^(-x))/(1+e^x)dx is equal to d) int1/(sqrt(1-e^(2x)))dx is equal to COLUMN II p) x-log[1+sqrt(1-e^(2x)]+c q) log(e^x+1)-x-e^(-x)+c r) log(e^(2x)+1)-x+c s) -1/(2(e^(2x)+1))+c

Column I, a) int(e^(2x)-1)/(e^(2x)+1)dx is equal to b) int1/((e^x+e^(-x))^2)dx is equal to c) int(e^(-x))/(1+e^x)dx is equal to d) int1/(sqrt(1-e^(2x)))dx is equal to COLUMN II p) x-log[1+sqrt(1-e^(2x)]+c q) log(e^x+1)-x-e^(-x)+c r) log(e^(2x)+1)-x+c s) -1/(2(e^(2x)+1))+c

Column I, a) int(e^(2x)-1)/(e^(2x)+1)dx is equal to b) int1/((e^x+e^(-x))^2)dx is equal to c) int(e^(-x))/(1+e^x)dx is equal to d) int1/(sqrt(1-e^(2x)))dx is equal to COLUMN II p) x-log[1+sqrt(1-e^(2x)]+c q) log(e^x+1)-x-e^(-x)+c r) log(e^(2x)+1)-x+c s) -1/(2(e^(2x)+1))+c

Column I, a) int(e^(2x)-1)/(e^(2x)+1)dx is equal to b) int1/((e^x+e^(-x))^2)dx is equal to c) int(e^(-x))/(1+e^x)dx is equal to d) int1/(sqrt(1-e^(2x)))dx is equal to COLUMN II p) x-log[1+sqrt(1-e^(2x)]+c q) log(e^x+1)-x-e^(-x)+c r) log(e^(2x)+1)-x+c s) -1/(2(e^(2x)+1))+c

int (1)/(e^(x)+2e^(-x)-3)dx=

int(e^x+1/(e^x))^2\ dx

(i) int (dx)/(e^x+e^(-x)) (ii) int e^x/(e^(2x)+1) dx

Evaluate: int(e^x)/((1+e^x)^2)\ dx