`int(sec^2(x))/(tan(x)(tan(x)+1))dx`

Let's apply integral substitution: `u=tan(x)`

`du=sec^2(x)dx`

`=int1/(u(u+1))du`

Now let's create the partial fraction template for the integrand,

`1/(u(u+1))=A/u+B/(u+1)`

Multiply the above equation by the denominator,

`1=A(u+1)+B(u)`

`1=Au+A+Bu`

`1=(A+B)u+A`

Equating the coefficients of the like terms,

`A+B=0` ------------(1)

`A=1`

Plug in the value of A in the equation 1,

`1+B=0`

`=>B=-1`

Plug back the values of A and B in the partial fraction template,

`1/(u(u+1))=1/u+(-1)/(u+1)`

`=1/u-1/(u+1)`

`int1/(u(u+1))du=int(1/u-1/(u+1))du`

Apply the sum rule,

`=int1/udu-int1/(u+1)du`

Use the common integral:`int1/xdx=ln|x|`

`=ln|u|-ln|u+1|`

Substitute back `u=tan(x)`

and add a constant C to the solution,

`=ln|tan(x)|-ln|tan(x)+1|+C`

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