Evaluate integral with Partial fraction expansion

Question

i'm having problem with my equation-system in this task.

$\int \frac {1-7x}{(x^2 + 1)(x-3)} \text{d}x$

Done this:

$\int \frac {Ax + B}{(x^2 +1)} + \frac {C}{(x-3)} \text{d}x $

And got this system: 1-7x = (A+C)x^2 - (3A + B)x - (3B + C)

A+C = 0

3A + B = -7

3B + C = 1

My problem is that i cant find a value from my system. Have i done something wrong?

Thanks

Answer 1

As has been stated: \begin{align} \frac{1-7x}{(x^{2}+1)(x-3)} = \frac{ax+b}{x^{2}+1} + \frac{c}{x-3} \end{align} which leads to $1 - 7x = (a+c)x^{2} + (b-3a)x + (c-3b)$ for which $a+c=0$, $b-3a = -7$, and $c-3b=1$. This leads to $a=-c$, and $b+3c=-7, c-3b=1$. Now $c=1+3b$ for which $-7 = b+3(1+3b) = 3 + 10b$ or $b=-1$, $c=-2$, and $a=2$. The fraction then becomes \begin{align} \frac{1-7x}{(x^{2}+1)(x-3)} = \frac{2x-1}{x^{2}+1} - \frac{2}{x-3}. \end{align} With the fractions being decomposed the integral in question then becomes \begin{align} I &= \int \frac{1-7x}{(x^{2}+1)(x-3)} \, dx = \int \left(\frac{2x-1}{x^{2}+1} - \frac{2}{x-3}\right) \, dx \\ &= \ln(x^{2} + 1) - \int \frac{dx}{x^{2}+1} - 2 \ln(x-3) \\ &= \ln(x^{2} + 1) - \tan^{-1}(x) - 2 \ln(x-3) + c_{0} \end{align}

Answer 2

You should have $$1-7x = (Ax+B)(x-3) + C(x^2 + 1) = (A + C)x^2 +(-3A + B) x + (-3B +C)$$

which gives you the system of equations:

$A+C = 0$

$-3A+B=-7$

$-3B+C = 1$

Answer 3

You started to decompose $$ \frac{1-7x}{(x^2+1)(x+3)} $$ (Note the + sign in ($x+3)$ instead of a minus.)

So both your $B$ coefficients in the second and third equations are minus what they should be.

Even so, your system could be solved: $$A+C = 0 \Rightarrow C = -A \\ 3A+B = -7 \Rightarrow B = -3A - 7 \\ 3(-3A - 7) + (-A) = 1 \Rightarrow A = -2.2 \Rightarrow B = -13.6, C=2.2 $$ The solution to the correct equations will be the same, except $B$ will be $+6.6$.