Indefinite Integral Confusion

Question

Solve the indefinite integral $$\int\frac{x^3−11x^2+x+2}{x^4−2x^3}\text{d}x.$$ My answer was $\frac{1}{2 x^2}+\frac{1}{x}-4\log(2-x)+5\log(x)+ C$ and I put that into my webwork and it was wrong. Then I tried again and got $5\log(x) + 4\log(x - 2)+ \frac{1}{x} + \frac{1}{2(x^2)}$ and that was still wrong. What am I doing wrong?

Answer 1

If the integral is evaluated over the real numbers, then you will want to use absolute values for the logarithmic terms: $$\frac{1}{2x^2} + \frac{1}{x} - 4 \log |2-x| + 5 \log |x| + C.$$ Also, if your web-based homework system is picky about distinguishing between $\log$ and $\ln$, you will want to use $\ln$. Mathematicians typically write $\log$ to mean $\ln$, because the base-10 logarithm $\log_{10} x$ is uncommon in pure mathematics. Some computer algebra systems agree with this convention, and by default, a logarithm is always a natural (base-$e$) logarithm unless otherwise specified.

Answer 2

We can write \begin{align} \frac{x^3-11x^2+x+2}{x^4-2x^3} & = \frac{A_1}{x}+\frac{A_2}{x^2}+\frac{A_3}{x^3}+\frac{B}{x-2} \end{align} for $A_1$, $A_2$, $A_3$ and $B$ real numbers such that \begin{align} x^3-11x^2+x+2 & =\left[\frac{A_1}{x}+\frac{A_2}{x^2}+\frac{A_3}{x^3}+\frac{B}{x-2}\right]x^3(x-2) \\ & = A_1x^2(x-2)+A_2x(x-2)+A_3(x-2)+Bx^3 \\ & = (A_1 + B)x^3+(-2A_1+A_2)x^2+(-2A_2+A_3)x-2A_3 \end{align} So \begin{array}{rcrcrcrcr} A_1 & & & & & + & B & = & 1 \\ -2A_1 & + & A_2 & & & & & = & -11 \\ & - & 2A_2 & + & A_3 & & & = & 1 \\ & & & - & 2A_3 & & & = & 2 \end{array} Solving this system of equations we obtain $A_1=5$, $A_2=-1$ $A_3=-1$ and $B=-4$. Then \begin{align} \int{\frac{x^3-11x^2+x+2}{x^4-2x^3}dx} & = \int{\frac{5}{x}dx}-\int{\frac{1}{x^2}dx}-\int{\frac{1}{x^3}dx}-\int{\frac{4}{x-2}dx} \\ & =5\log|x|+\frac{1}{x}+\frac{1}{2x^2}-4\log|x-2|+C \end{align} Where $C$ is an integration constant.