`I=int(n)/(sqrt(n+4))dn`

int(n)/(sqrt(n-3))dn

I=int(n^2dn)/(sqrt(a^(3)-n^(3)))

Let I_(n)=int(x^(n))/(sqrt(x^(2)+2x+5))dx then I_(n)=(x^(m-1))/(n)sqrt(x^(2)+2x+5)-lambda I_(n-1)-mu I_(n-2) where lambda,mu and m' are involving n' then

int((a+sqrt(x))^(n))/(sqrt(x))dx,n!=-1

"int((a+sqrt(x))^(n))/(sqrt(x))dx,n!=-1

If int(1)/(sqrt(e^(4x)-36))dx=m.tan^(-1)[n.sqrt(e^(4x)-36)]+c , then : (m, n)-=

int sqrt((1-n)/(1+n))dn

int sqrt(a^(2)-n^(2))dn

If I_(n)=int(x^(n)dx)/(sqrt(x^(2)+a)) then prove that I_(n)+(n-1)/(n)al_(n-2)=(1)/(n)x^(n-1)*sqrt(x^(2)+a)

int(x^(n-1))/(sqrt(1+4x^(n)))dx