`int_0^1(x^(1//3)+1)^2dx`

Find int_0^1 log(2+x^(1/3))/x^(1/3)dx

Find int_0^1 log(2+x^(1/3))/x^(1/3)dx

int_0^1(dx)/(3x+2)

int_0^1(x^2)/(1+x^2)dx=?

int_0^1 x^2(1-x)^(3/2) dx=

Let I_1=int_0^1 (dx)/(1+x^(1/3)) and I_2=int_0^1 (dx)/(1+x^(1/4)) then 4I_1+3I_2=

If mgt0, ngt0 , the definite integral I=int_0^1 x^(m-1)(1-x)^(n-1)dx depends upon the values of m and n is denoted by beta(m,n) , called the beta function.For example, int_0^1 x^3 (1-x)^4dx=int_0^1 x^(4-1) (1-x)^(5-1) dx=beta(4,5) and int_0^1 x^(3/2) (1-x)^((-1)/2)dx=int_0^1 x^(5/2-1) (1-x)^(1/2-1)dx=beta(5/2,1/2) .Obviously, beta(n,m)=beta(m,n) .Now answer the question:If int_0^2 (8-x^3)^((-1)/3)dx=kbeta(1/3,2/3) , then k equals to (A) 1 (B) 1/2 (C) 1/3 (D) 1/4

int_-a^a(1+x^3)^(-1)dx= (A) 0 (B) 2int_0^a(1-x^6)^(-1)dx (C) 2int_0^a(1+x^3)^(-1)dx (D) 2int_0^a[1+(a-x^3)]^(-1)dx

int_-a^a(1+x^3)^(-1)dx= (A) 0 (B) 2int_0^a(1-x^6)^(-1)dx (C) 2int_0^a(1+x^3)^(-1)dx (D) 2int_0^a[1+(a-x^3)]^(-1)dx