If `I(m,n)=int_0^1 t^m(1+t)^n.dt`, then the expression for `I(m,n)` in terms of `I(m+1,n-1)` is:

If I(m,n)=int_0^1x^(m-1)(1-x)^(n-1)dx , then

If I_(m,n)= int_(0)^(1) x^(m) (ln x)^(n) dx then I_(m,n) is also equal to

If L(m,n)=int_(0)^(1)t^(m)(1+t)^(n),dt , then prove that L(m,n)=(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)

If log_(a)m = n , express a^(n-1) in terms of a and m.

If (a^(m))^(n)=a^(m^(n)) , then express 'm' in the terms of n is (agt0, ane0, mgt1, ngt1)

I_n = int_0^(pi/4) tan^n x dx , then the value of n(l_(n-1) + I_(n+1)) is

IfI(m , n)=int_0^1x^(m-1)(1-x)^(n-1)dx ,(m , n in I ,m ,ngeq0),t h e n (a) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m-n))dx (b) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m+n))dx (c) I(m , n)=int_0^oo(x^(n-1))/((1+x)^(m+n))dx (d) I(m , n)=int_0^oo(x^n)/((1+x)^(m+n))dx

Let m,n be two positive real numbers and define f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx and g(m,n)=int_(0)^(1)x^(m-1)(1-x)^(n-1)dx . It is known that f(n) for n gt 0 is finite and g(m, n) = g(n, m) for m, n gt 0 . int_(0)^(1)(x^(m-1)+x^(n-1))/((1+x)^(m+n))dx=

Let m,n be two positive real numbers and define f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx and g(m,n)=int_(0)^(1)x^(m-1)(1-m)^(n-1)dx . It is known that f(n) for n gt 0 is finite and g(m, n) = g(n, m) for m, n gt 0. int_(0)^(1)x^(m)(log_(e).(1)/(x))dx=

If I(m) = int_0^pi ln(1-2m cos x + m^2)dx , then I(1)=