`int(sec^2(x))/(tan^2(x)+5tan(x)+6)dx`

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

`=>du=sec^2(x)dx`

`=int1/(u^2+5u+6)du`

Now we have to write down integrand as sum of partial fraction function, but first we will have to factor the denominator,

`1/(u^2+5u+6)=1/(u^2+2u+3u+6)`

`=1/(u(u+2)+3(u+2))`

`=1/((u+2)(u+3))`

Now let's create partial fraction template,

`1/((u+2)(u+3))=A/(u+2)+B/(u+3)`

Multiply the above equation by the denominator,

`=>1=A(u+3)+B(u+2)`

`1=Au+3A+Bu+2B`

`1=(A+B)u+3A+2B`

...

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`int(sec^2(x))/(tan^2(x)+5tan(x)+6)dx`

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

`=>du=sec^2(x)dx`

`=int1/(u^2+5u+6)du`

Now we have to write down integrand as sum of partial fraction function, but first we will have to factor the denominator,

`1/(u^2+5u+6)=1/(u^2+2u+3u+6)`

`=1/(u(u+2)+3(u+2))`

`=1/((u+2)(u+3))`

Now let's create partial fraction template,

`1/((u+2)(u+3))=A/(u+2)+B/(u+3)`

Multiply the above equation by the denominator,

`=>1=A(u+3)+B(u+2)`

`1=Au+3A+Bu+2B`

`1=(A+B)u+3A+2B`

Equating the coefficients of the like terms,

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

`3A+2B=1` -----------------(2)

From equation 1:`A=-B`

Substitute A in equation 2,

`3(-B)+2B=1`

`-3B+2B=1`

`=>B=-1`

Plug in the values in the partial fraction template,

`1/((u+2)(u+3))=1/(u+2)-1/(u+3)`

`int1/(u^2+5u+6)du=int(1/(u+2)-1/(u+3))du`

Apply the sum rule,

`=int1/(u+2)du-int1/(u+3)du`

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

`=ln|u+2|-ln|u+3|`

Substitute back `u=tan(x)`

and add a constant C to the solution,

`=ln|tan(x)+2|-ln|tan(x)+3|+C`