# evaluate the integral: ∫(3x^2+3x+3)/(x^2+1)dx

mvcdc | Student, Graduate | (Level 2) Associate Educator

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*Note that log here is the natural logarithm - by log(u) i mean ln(u).

mvcdc | Student, Graduate | (Level 2) Associate Educator

Posted on

First, we simpify the integrand:

`(3x^2 + 3x + 3)/(x^2 + 1) = 3 (x^2 + x + 1)/(x^2 + 1)`

We then perform long division to get:

`3 (1 + x/(x^2 + 1))`

The integral that we want to solve now is:

`int [ 3 (1 + x/(x^2 + 1) ) ] dx`

`3 int [ 1 + x/(x^2 + 1) ] dx`

`3 int 1 dx + 3 int x/(x^2 + 1) dx`

`3 int dx + 3 int x/(x^2 + 1) dx`

Using the substitution method, we let `u = x^2 + 1` , so that `du = 2xdx` and `xdx = (du)/2` . We now have:` `` `

`3 int dx + 3 int 1/(2u) du`

`3 int dx + 3/2 int (1/u) du`

The integrals are now easier to solve: `3int dx = 3x + C_1` and `3/2 int 1/u du = 3/2 log(u) +C_2` . Together, we get the following:

`3x + 3/2 log(u) + C` where C is the integration constant.

Plugging back `u = x^2 + 1` . We get:

`int (3x^2+3x+3)/(x^2+1)dx = 3x + 3/2 log(x^2 + 1) + C`