`int (4x)/(x^3 + x^2 + x + 1) dx` Evaluate the integral

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lemjay eNotes educator| Certified Educator

`int(4x)/(x^3+x^2+x+1)dx`

To solve, apply the partial fraction decomposition.  

To do so, factor the denominator.

`int(4x)/(x^3+x^2+x+1)dx = int(4x)/((x+1)(x^2+1))dx`

Then, express the integrand as sum of proper rational expressions.

`(4x)/((x+1)(x^2+1))=A/(x+1)+(Bx+C)/(x^2+1)`

Multiply both sides by the LCD.

`4x =A(x^2+1)+(Bx+C)(x+1)`

`4x = Ax^2+A + Bx^2+Bx+Cx+C`

`4x=(A+B)x^2+(B+C)x + A+C`

Express the left side as a polynomial with degree 2.

`0x^2+4x+0=(A+B)x^2+Cx+A+C`

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

x^2:

`0=A+B`     (Let this be EQ1.)

x:

`4=B+C`     (Let this be EQ2.)    

Constant:

`0=A+C`    (Let this be EQ3.)

To solve for the values of A, B and C, isolate the A in EQ1 and the C in EQ2.

EQ1:    

 `0=A+B`

`-B=A`

EQ2:  

`4 = B + C`

`4 - B = C`

Plug-in them to EQ3.

EQ3:

`0=A+C`

`0=-B+4-B`

`0=-2B+4`

`-4=-2B`

`2=B`

And, plug-in the value of B to EQ1 and EQ2.

EQ1:

`0 =A + B`

`0=A+2`

`-2=A`

EQ2:

`4=B+C`

`4=2+C`

`2=C`

So the partial fraction decomposition of the integrand is:

`(4x)/(x^3+x^2+x+1) = -2/(x+1) + (2x+2)/(x^2+1)=-2/(x+1)+(2x)/(x^2+1)+2/(x^2+1)`

Then, take the integral of it.

`int (4x)/(x^3+x^2+x+1)dx`

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

`=int -2/(x+1)dx + int(2x)/(x^2+1)dx + int2/(x^2+1)dx`

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

 

Therefore, `int (4x)/(x^3+x^2+x+1)dx=-2ln|x+1| + ln|x^2+1|+2tan^(-1)(x) +C` .         

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