`int n^3/(5-2n^4)^3`dn

int nsec^2 n dn

If A={p:p=((n+2)(2n^(5)+3n^(4)+4n^(3)+5n^(2)+6))/(n^(2)+2n), n p in Z^(+)} then the number of elements in the set A, is

Find the value of n in each of the following: (2^2)^n=(2^3)^4 (ii) 2^(5n)-:2^n=2^4 (iii) 2^n^(-5)xx5^n^(-4)=5

Integrate int((x^(2n)-x^(2n-3)))/((2x^(3)+1)^(n))dx .

find the value of n, given : (4^(n-2)xx2^(n-5)xx6xx2^(n+3))/(3xx6^(2))=(4)/(18)

If sum_(k=1)^(n) (sum_(m=1)^(k) m^(2))=an^(4)+bn^(3)+cn^(2)+dn+e , then

Find the value of n in each of the following: (i) 2^(n-7)xx5^(n-4)=1250 (ii) 5^(n-2)xx3^(2n-3)=135

The value of lim_(n to oo) (2n^(2) - 3n + 1)/(5n^(2) + 4n + 2) equals

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

Simplify: find the value of n (n)/(2)-(3n)/(4)+(5n)/(6)=21