How do you evaluate the integral $int (x^3+x^2+x+1)/(x(x+4))$ ?

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Answer 1 · 2017-01-03T20:57:30

$1/2x^2 - 3x + 1/4ln|x| + 51/4ln|x + 4| + C$

We start by dividing $x^3 + x^2 + x + 1$ by $x^2 + 4x$ .

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So, we have $(x^3 + x^2 + x + 1)/(x^2 + 4x) = x - 3 + (13x+ 1)/(x^2 + 4x)$

We can integrate $x - 3$ using the power rule. However, we will need to use partial fractions for the second part of the integral.

$x^2 + 4x$ can be factored as $x(x + 4)$ .

$A/x + B/(x + 4) = (13x + 1)/(x(x + 4))$

$A(x + 4) + B(x) = 13x + 1$

$Ax + 4A + Bx = 13x + 1$

$(A + B)x + 4A = 13x + 1$

Write a system of equations:

${(A + B = 13), (4A = 1):}$

Solving, we get $A = 1/4$ and $B = 51/4$ .

$=intx - 3 + 1/(4x) + 51/(4(x + 4)) dx$

$= 1/2x^2 - 3x + 1/4ln|x| + 51/4ln|x + 4| + C$

Hopefully this helps!

How do you evaluate the integral int (x^3+x^2+x+1)/(x(x+4))int (x^3+x^2+x+1)/(x(x+4))?