You need to use the linearity property of integrals such that:

`int (x+6)/(x^2+36)dx = int x/(x^2+36)dx + 6 int 1/(x^2+36)dx`

You should use substitution to solve `int x/(x^2+36)dx` such that:

`x^2+36 = t => 2xdx = dt => xdx = (dt)/2`

`int x/(x^2+36)dx = int ((dt)/2)/t`

`int ((dt)/2)/t = (1/2)ln|t|...

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You need to use the linearity property of integrals such that:

`int (x+6)/(x^2+36)dx = int x/(x^2+36)dx + 6 int 1/(x^2+36)dx`

You should use substitution to solve `int x/(x^2+36)dx` such that:

`x^2+36 = t => 2xdx = dt => xdx = (dt)/2`

`int x/(x^2+36)dx = int ((dt)/2)/t`

`int ((dt)/2)/t = (1/2)ln|t| + c`

Substituting back `x^2+36` for t yields:

`int x/(x^2+36)dx = (1/2)ln(x^2+36) + c`

You need to solve the integral `int 1/(x^2+36)dx` using the following formula such that:

`int 1/(x^2+a^2)dx = (1/a)arctan(x/a) + c`

Reasoning by analogy yields:

`int 1/(x^2+36)dx =(1/6)arctan(x/6) + c `

`int (x+6)/(x^2+36)dx = (1/2)ln(x^2+36) + 6*(1/6)arctan(x/6) + c `

`int (x+6)/(x^2+36)dx = ln sqrt(x^2+36) + arctan(x/6) + c`

**Hence, evaluating the given indefinite integral yields`int (x+6)/(x^2+36)dx = ln sqrt(x^2+36) + arctan(x/6) + c.` **