Since the problem provides the information that `int_(1/x)^x f(t) dt = int_1^x t^3f(t) dt` , yields:

`int_1^x t^3f(t) = int_1^x t^3/(1+t^2)(1+t^3) dt`

You should add and subtract 1 such that:

`int_1^x t^3 (t^3+1-1)/(1+t^2)(1+t^3) dt`

Using the linearity of integral yields:

`int_1^x (t^3+1-1)/(1+t^2)(1+t^3) dt = int_1^x (t^3+1)/(1+t^2)(1+t^3) dt - int_1^x (1)/(1+t^2)(1+t^3) dt`

`int_1^x (t^3+1-1)/(1+t^2)(1+t^3) dt = int_1^x 1/(1+t^2) dt- int_1^x (1)/(1+t^2)(1+t^3) dt`

You should use partial fraction decomposition to solve `int_1^x (1)/(1+t^2)(1+t^3) dt` such that:

`1/(1+t^2)(1+t^3) = (At + B)/(t^2+1) + C/(t+1) + (Dt+E)/(t^2-t+1)`

`1 = (At + B)(t^3+1) + C(t^2+1)(t^2-t+1) + (Dt+E)(t^2+1)(t+1)`

`1 = At^4 + At + Bt^3 + B + Ct^4 - Ct^3 + Ct^2 + Ct^2 - Ct + C + (Dt^3 + Dt + Et^2 + Et)(t+1)`

`1 = At^4 + At + Bt^3 + B + Ct^4 - Ct^3 + Ct^2 + Ct^2 - Ct + C + Dt^4 + Dt^2 + Et^3 + Et^2 + Dt^3 + Dt + Et^2 + Et`

`1 = t^4(A+C+D) + t^3(B - C + E + D) + t^2(2C + D + 2E) + t(A - C + D + E) + B + C`

Equating the coefficients of like powers yields:

`A+C+D = 0`

`B - C + E + D = 0`

`2C + D + 2E = 0`

`A - C + D + E = 0`

`B + C = 1`

`2B+E+D = 1 => 2B+E+D-2A-2D-E = 1 => 2B-2A-D = 1`

`2A+2D+E=0`

`A = 0 => B = 1/4 => C = 1 - 1/4 = 3/4 => D = -3/4=> E = 3/8`

`1/(1+t^2)(1+t^3) = 1/(4(t^2+1)) + 3/(4(t+1))- (3/4)(2t-1)/(t^2-t+1)`

`int_1^x (1)/(1+t^2)(1+t^3) dt = int_1^x 1/(4(t^2+1))dt+ int_1^x 3/(4(t+1)) dt- (3/4) int_1^x (2t-1)/(t^2-t+1) dt`

Using the fundamental theorem of calculus yields:

`int_1^x (1)/(1+t^2)(1+t^3) dt = ((1/4) arctan(t^2+1) + (3/4)ln|t+1| - (3/4)ln(t^2-t+1))|_1^x`

`int_1^x (1)/(1+t^2)(1+t^3) dt = (1/4) (arctan(x^2+1) - arctan 2) + (3/4)(ln|x+1| - ln2) - (3/4)(ln(x^2-x+1))`

`int_1^x (t^3+1-1)/(1+t^2)(1+t^3) dt = (3/4)(arctan(x^2+1) - arctan 2) - (3/4)(ln|x+1| - ln2)+ (3/4)(ln(x^2-x+1))`

**Hence, evaluating the given integral yields `int_1^x (t^3+1-1)/(1+t^2)(1+t^3) dt = (3/4)(arctan(x^2+1) - arctan 2) - (3/4)(ln|x+1| - ln2) + (3/4)(ln(x^2-x+1)).` **