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int(x^3-x^2+x-1)/(x-1)dx[N C E R T...

`int(x^3-x^2+x-1)/(x-1)dx[N C E R T`

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int (e^(3x)+1)^2 e^(3x)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

The integral int(1+x-(1)/(x))e^(x+(1)/(x))dx is equal to (1)(x-1)e^(x+(1)/(x))+C(2)xe^(x+(1)/(x))+C(3)(x+1)e^(x+(1)/(x))+C(2)-xe^(x+(1)/(x))+C

For int (x -1)/( (x +1 ) ^(3)) e ^(x) dx = e ^(x) f (x) + c , f (x) =(x +1) ^(2).

Let f(x)=(e^(x)+1)/(e^(-x)-1) and int_(0)^(1)x^(3)(e^(x)+1)/(e^(x)-1) dx=alpha"then", int_(-1)^(1)int t^(3)f(t)dt is equal to

int( x-1)e^(-x) dx is equal to : a) ( x- 2)e^(x) + C b) x e^(-x) + C c) - x e^(x) + C d) ( x+1)e^(-x) +C