`L e tA=int_0^oo(logx)/(1+x^3)dxdotT h e nfin dt h ev a l u eofint_0^oo(xlogx)/(1+x^3)dx`in terms of `Adot`

`L e tA=int_0^oo(logx)/(1+x^3)dxdotT h e nfin dt h ev a l u eofint_0^o

1.	`L e tA=int_0^oo(logx)/(1+x^3)dxdotT h e nfin dt h ev a l u eofint_0^oo(xlogx)/(1+x^3)dx`in terms of `Adot`
Answer» `B=int_(0)^(oo)(x log x)/(1+x^(3))dx` `=int_(0)^(oo) ((x+1)logx-log)/(1+x^(3))dx` `=int_(0)^(oo) (logx)/(x^(2)-x+1)dx-A` let `I=int_(0)^(oo) (logx)/(x^(2)-x+1) dx` Put `x=1/t` `:. I=int_(oo)^(0) ("log"1/t)/(-1/(t^(2))-1/t+1)((-dt)/(t^(2)))` `=-int_(0)^(oo) (logt)/(t^(2)-t+1)dt` `=-I` or `2I=0` or `I=0` `:.B=-A`

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
The area bounded by the curves `y=cosx` and `y=sinx` between the ordinates `x=0` and `x=(3pi)/2` isA. `4sqrt(2) + 2`B. `4sqrt(2) +1`C. `4sqrt(2) + 1`D. `4sqrt(2) - 2`
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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
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Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`