# Integration by partial fractions, #2

1. Jan 22, 2006

### Xcron

This next problem is rather strange and it once again involves quadratic factors and I am not able to get the correct answer.

The problem is:
$$\int \frac{7x^3-3x^2+73x+53}{(x-1)^2(x^2+25)}dx$$

Step I:
$$7x^3-3x^2+73x+53 = A(x-1)(x^2+25)+B(x^2+25)+(Cx+D)(x-1)^2$$

I easily get the value of B by plugging in 1 for x and it turns out to be 5. After that I organize all of the degrees and the right side and use the method of equating coefficients, but that turns out to be an extremely tedious monster which leads me to the wrong values for coefficients.

2. Jan 23, 2006

### benorin

Just because x=1 is the only real root of the denominator, doesn't mean you can't just plug in other values of x to solve for other variables, e.g. use x=0 or x=431 or x=whatever. If you know how to use complex numbers, you should plug in x=5i and x=-5i.

3. Jan 23, 2006

### benorin

A=3, B=5, C=4, D=3

and thus

$$\int \frac{7x^3-3x^2+73x+53}{(x-1)^2(x^2+25)}dx = -\frac{5}{x-1}+3\ln (x-1) + 2\ln (x^2+25) +\frac{3}{5}\mbox{arctan} \left( \frac{x}{5}\right)+C$$

I used Maple.