# `int (x^3 + x^2 + 2x + 1)/((x^2 + 1)(x^2 + 2)) dx` Evaluate the integral

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

To solve, apply partial fractions decomposition.

To express the integrand as sum of proper rational expressions, set the equation as follows:

`(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (Ax+B)/(x^2+1)+(Cx+D)/(x^2+2)`

Multiply both sides by the LCD.

`x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)`

`x^3+x^2+2x+1=Ax^3+2Ax+Bx^2+2B + Cx^3+Cx+Dx^2+D`

`x^3+x^2+2x+1=(A+C)x^3+(B+D)x^2+(2A+C)x+2B+D`

For the two sides to be equal, the two polynomials should be the...

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

To solve, apply partial fractions decomposition.

To express the integrand as sum of proper rational expressions, set the equation as follows:

`(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (Ax+B)/(x^2+1)+(Cx+D)/(x^2+2)`

Multiply both sides by the LCD.

`x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)`

`x^3+x^2+2x+1=Ax^3+2Ax+Bx^2+2B + Cx^3+Cx+Dx^2+D`

`x^3+x^2+2x+1=(A+C)x^3+(B+D)x^2+(2A+C)x+2B+D`

For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.

x^3:

`1=A+C `   (Let this be EQ1.)

x^2:

`1=B+D`     (Let this be EQ2.)

x:

`2=2A+C`     (Let this be EQ3.)

Constant:

`1=2B+D`     (Let this be EQ4.)

To solve for the values of A, B, C and D, isolate the C in EQ1.

`1=A+C`

`1-A=C`

Plug-in this to EQ3.

`2=2A+C`

`2=2A+1-A`

`2=A+1`

`1=A`

Plug-in the value of A to EQ1.

`1=A+C`

`1=1+C`

`0=C`

Also, isolate the D in EQ2.

`1=B+D`

`1-B=D`

Plug-in this to EQ4.

`1=2B+D`

`1=2B+1-B`

`1=B+1`

`0=B`

And plug-in the value of B to EQ2.

`1=B+D`

`1=0+D`

`1=D`

So the partial fraction decomposition of the integrand is:

`(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (1x+0)/(x^2+1)+(0x+1)/(x^2+2)=x/(x^2+1)+1/(x^2+2) `

Taking the integral of this result to:

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

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

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

For the first integral, apply u-substitution method.

`u=x^2+1`

`du=2xdx`

`(du)/2=xdx`

`= int 1/u*(du/2) + int 1/(x^2+2)dx`

`=1/2 int 1/u du + int 1/(x^2+2)dx`

`= 1/2ln|u| + 1/sqrt2 tan^(-1) (x/sqrt2)+C`

And, substitute back `u = x^2+1` .

`=1/2ln |x^2+1|+1/sqrt2tan^(-1)(x/sqrt2)+C`

Therefore, `int (x^3+x^2+2x+1)/((x^2+1)(x^2+2))dx=1/2ln |x^2+1|+1/sqrt2tan^(-1)(x/sqrt2)+C` .

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