# Integral x-sqrt(arctg x)/1+x^2 dx?

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### 1 Answer

You should use first the property of linearity of integral, such that:

`int (x - sqrt(arctan x))/(1 + x^2) dx = int x/(1 + x^2) dx - int (sqrt(arctan x))/(1 + x^2) dx`

You should use substitution method to solve both integrals, hence, you need to come up with the following substitution for the first integral, such that:

`x^2 + 1 = t => 2x dx = dt => xdx = (dt)/2`

Changing the variable yields:

`int x/(1 + x^2) dx = int ((dt)/2)/t`

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

Replacing back `x^2 + 1` for `t` yields:

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

You need to come up with the following substitution for the second integral, such that:

`arctan x = y => 1/(1 + x^2) dx = dy`

Replacing the variable yields:

`int (sqrty) dy = int y^(1/2) dy = (y^(1/2+1))/(3/2) + c`

`int (sqrty) dy = (2/3)y*sqrt y + c`

Replacing back `arctan x` for `y` yields:

`int (sqrt(arctan x))/(1 + x^2) dx = (2/3)arctan x*sqrt (arctan x) + c`

**Hence, evaluating the given indefinite integral yields `int (x - sqrt(arctan x))/(1 + x^2) dx = (1/2)ln(x^2 + 1) + (2/3)arctan x*sqrt (arctan x) + c.` **