How do you integrate (x^2 + 33)/(x^3 + x^2)x2+33x3+x2 using partial fractions?

2 Answers
Nov 13, 2016

y = 34ln|x + 1| - 33ln|x| - 33/x + Cy=34ln|x+1|33ln|x|33x+C

Explanation:

x^3 + x^2 = x^2(x + 1)x3+x2=x2(x+1)

So,

(Ax + B)/x^2 + C/(x+ 1) = (x^2 + 33)/((x^2)(x + 1))Ax+Bx2+Cx+1=x2+33(x2)(x+1)

(Ax + B)(x + 1) + Cx^2 = x^2 + 33(Ax+B)(x+1)+Cx2=x2+33

Ax^2 + Bx + Ax +B + Cx^2 = x^2 + 33Ax2+Bx+Ax+B+Cx2=x2+33

(A + C)x^2 + (A + B)x + B = x^2 + 33(A+C)x2+(A+B)x+B=x2+33

So,

{(A + C = 1), (A + B = 0), (B = 33):}

Solving, we get that A = -33, B = 33, C = 34.

Hence, the partial fraction decomposition is as follows:

(-33x + 33)/x^2 + 34/(x + 1)

We can integrate the second term as 34ln|x + 1| + C. However, we can decompose the first term further.

int(33- 33x)/x^2dx = int(33(1 - x))/x^2dx = 33int(1- x)/x^2dx

A/x + B/x^2 = (1 - x)/x^2

Ax + B = 1 - x

Here, we have that A = -1 and B = 1.

Hence,

=>33int(1/x^2 - 1/x)dx

=>33int(x^(-2) - 1/x)dx

=>33(-1/x - ln|x|) + C

Putting this and the part that we integrated above together gives

=>-33/x - 33ln|x| + 34ln|x + 1| + C

Hopefully this helps!

See answer below:
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Explanation:

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