Integrate the indefinite integral. \int (3x^(3)-x^(2)+6x-4)/((x^(2)+1)(x^(2)+2))dx

First, we should represent the function under the integral as a sum of partial fractions. Then it will integrate easily using arctangent and/or logarithm.

The decomposition should exist in the form

( 3x^3 - x^2 + 6x - 4 ) / ( ( x^2 + 1 ) ( x^2 + 2 ) ) = ( Ax + B ) / ( x^2 + 1 ) + ( Cx + D ) / ( x^2 + 2 ) .

To find the constants, multiply by the denominator:

3x^3 - x^2 + 6x - 4 = ( Ax + B ) ( x^2 + 2 ) + ( Cx + D ) ( x^2 + 1 ) ,  i.e.

3x^3 - x^2 + 6x - 4 = ( A + C ) x^3 + ( B + D ) x^2 + ( 2A + C ) x + ( 2B + D ) .

This gives the system of equations

A + C = 3 ,  B + D = -1 ,  2A + C = 6 ,  2B + D = -4,

which clearly has only one solution A = 3 ,  C = 0 ,  B = -3 ,  D = 2 .

Then the indefinite integral is equal to

3 int x / (x^2+1) dx - 3 int (dx) / (x^2+1) dx + 2 int (dx)/(x^2+2) ,

which is simple and is equal to

3/2 ln(x^2+1) - 3 arctan (x) + sqrt(2) arctan(x/sqrt(2)) + C .

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