Compute the integrals:`int_0^oof(x^n+x^(-n))logx(dx)/x`

1.	Compute the integrals:`int_0^oof(x^n+x^(-n))logx(dx)/x`
Answer» Here limits (reciprocal) and type of functions (reciprocal terms are present i.e. `x` and `1//x`) suggest that we must substitute `1//t` for `x`. Let `t=1//x` or `x=1//t`. So `dx=-1/(t^(2))dt`. Also when `xto 0, t to oo` when `xto oo, t to 0`. Thus, `I=int_(0)^(oo) f(x^(n)+x^(-n)) In x(dx)/x` `=int_(oo)^(0)f(t^(-n)+t^(n))In(1/t)(-(dt)/(t^(2)))/(1/t)` `=-int_(0)^(oo) ft^(n)+t^(-n)In(t)(dt)/t` `=-1` or `2I=0` or `I=0`

Discussion

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.
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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4
`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
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))`