`int _(0)^(1)1/(2x-3 )dx`

I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx , I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1)dx . Then

Evaluate : int_(0)^(1)a^(2-3x)dx , as limit of sums .

The value of int_(0)^(1)({2x}-1)({3x}-1)dx , (where {} denotes fractional part opf x) is equal to :

The value of int _(0)^(1) ((x ^(6) -x ^(3)))/((2x ^(3)+1)^(3)) dx is equal to :

For x in(0,1) arrange f_(1)(x) = (1)/(9-x^(2)), f_(2)(x) = (1)/(9-2x^(2)) and f_(3)(x) = (1)/(9-x^(2)-x^(3)) in ascending order and hence prove that 1/6 ln2 lt int_(0)^(1)(1)/(9-x^(2)-x^(3)) dx lt (1)/(6sqrt(2)) ln 5 .

int_(0)^(1) |5x -3 | dx =?

int_(0)^(1) (1)/(x^(2)+x+1)dx

If I_(1)=int_(0)^(2pi)sin^(3)xdx and I_(2)=int_(0)^(1)ln((1)/(x)-1)dx , then

Evaluate :- int_(0)^(1)(3x^(2)+4)dx

int_(0)^(1) (1)/(x^(2) +2x+3)dx