What is the integral of (3x+1)dx/(2+x^2)?

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You should use first linearity of integral such that:

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

You need to use substitution to solve the integral `int (3x)/(x^2+2)dx ` such that:

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

`int (3x)/(x^2+2)dx = 3 int (dt)/2/t `

`3 int (dt)/2/t = (3/2) int (dt)/t => (3/2) int (dt)/t = (3/2)ln|t| + c`

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

`int (3x)/(x^2+2)dx = (3/2)ln(x^2 + 2) + c`

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

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

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

Reasoning by analogy yields:

`int 1/(x^2+2)dx = (1/sqrt2)tan^(-1)(x/sqrt2) + c`

`int (3x+1)dx/(2+x^2) = (3/2)ln(x^2 + 2) +(1/sqrt2)tan^(-1)(x/sqrt2) + c `

Hence, evaluating the given integral yields `int (3x+1)dx/(2+x^2) = (3/2)ln(x^2 + 2) + (1/sqrt2)tan^(-1)(x/sqrt2) + c.`

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