Answer to Question #184748 in Calculus for Angelo

Question

Evaluate the Integration of Rational Functions by Partial Fractions:

1. ∫ (x² +x-1/x³-x) dx

2. ∫ (x² -x+1/(x+1)³) dx

3. ∫ (x²+4x+10/x³ +2x² +5x) dx

4. ∫ (x²/(x-1)(x⁴ +8x² +16)) dx

Answer 1

Solution.

1)

$$\int \frac{x^2+x-1}{x^3-x}dx$$

$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}=\frac{Ax^2-A+Bx^2+Bx+Cx^2-Cx}{x(x-1)(x+1)}=\newline =\frac{x^2+x-1}{x^3-x}.$

From here

$\begin{matrix} A+B+C=1,\\ B-C=1,\\ -A=-1. \end{matrix}$

$A=1, B=\frac{1}{2}, C=-\frac{1}{2}.$

We will have

$\int \frac{x^2+x-1}{x^3-x}dx=-\frac{1}{2}\int \frac{1}{x+1}dx+\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx=\newline =-\frac{1}{2}\ln|x+1|+\ln|x|+\frac{1}{2}\ln|x-1|+C=\newline =\frac{\ln|x-1|-\ln|x+1|}{2}+\ln|x|+C,\newline \text{where C is some constant.}$

2)

$$\int \frac{x^2-x+1}{(x+1)^3}dx$$

$\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}=\frac{Ax^2+2Ax+BA+Bx+B+C}{(x+1)^3}=\newline =\frac{x^2-x+1}{(x+1)^3}.$

From here

$\begin{matrix} A=1,\\ 2A+B=-1,\\ A+B+C=1. \end{matrix}$

$A=1, B=-3, C=3.$

We will have

$\int \frac{x^2-x+1}{(x+1)^3}dx=\int \frac{1}{x+1}dx-3\int \frac{1}{(x+1)^2}dx+3\int \frac{1}{(x+1)^3}dx=\newline =\ln|x+1|+\frac{3}{x+1}-\frac{3}{2(x+1)^2}+C=\newline =\ln|x+1|+\frac{6x+3}{2x^2+4x+2}+C,\newline \text{where C is some constant.}$

3)

$$\int \frac{x^2+4x+10}{x^3+2x^2+5x}dx$$

$\frac{A}{x}+\frac{Bx+C}{x^2+2x+5}=\frac{Ax^2+2Ax+5A+Bx^2+Cx}{x(x^2+2x+5)}=\newline =\frac{x^2+4x+10}{x^3+2x^2+5x}.$

From here

$\begin{matrix} A+B=1,\\ 2A+C=4,\\ 5A=10. \end{matrix}$

$A=2, B=-1, C=0.$

We will have

$\int \frac{x^2+4x+10}{x^3+2x^2+5x}dx=2\int \frac{1}{x}dx-\int \frac{x}{x^2+2x+5}dx=\newline 2\int \frac{1}{x}dx-\int \frac{d(x^2+2x+5)}{x^2+2x+5}+\int \frac{dx}{x^2+2x+5}=2\ln|x|-\frac{1}{2}\ln|x^2+2x+5|+\frac{1}{2}\arctan{\frac{x+1}{2}}+C,\newline \text{where C is some constant.}$

4)

$$\int \frac{x^2}{(x-1)(x^4+8x^2+16)}dx$$

$\frac{A}{x-1}+\frac{Bx+C}{x^2+4}+\frac{Dx+E}{(x^2+4)^2}=\frac{Ax^4+8Ax^2+16A+Bx^4+4Bx^2-Bx^3-4Bx+Cx^3+4Cx-Cx^2-4C+Dx^2-Dx+Ex-E}{(x-1)(x^2+4)^2}=\newline =\frac{x^2}{(x-1)(x^2+4)^2}.$

From here

$\begin{matrix} A+B=0,\\ -B+C=0,\\ 8A+4B-C+D=1,\\ -4B+4C-D+E=0,\\ 16A-4C-E=0. \end{matrix}$

$A=\frac{1}{25}, B=-\frac{1}{25}, C=-\frac{1}{25}, \newline D=\frac{4}{5}, E=\frac{4}{5}.$

We will have

$\int \frac{x^2}{(x-1)(x^4+8x^2+16}dx=\frac{1}{25}\int \frac{1}{x-1}dx- \frac{1}{25}(\int \frac{x}{x^2+4}dx+\int \frac{1}{x^2+4}dx)+ \frac{4}{5}(\int \frac{x}{(x^2+4)^2}dx+\int \frac{1}{(x^2+4)^2}dx)= \frac{1}{25}\ln|x-1|- \frac{1}{25}(\frac{\ln(x^2+4)}{2}+\frac{\arctan \frac{x}{2}}{2})+ \frac{4}{5}(-\frac{1}{2(x^2+4)}+\frac{\arctan \frac{x}{2}}{16}+\frac{x}{8(x^2+4)})+C= \frac{\ln|x-1|}{25}-\frac{\ln(x^2+4)}{50}+\frac{3\arctan \frac{x}{2}}{100}-\frac{2}{5(x^2+4)}+\frac{x}{10(x^2+4)}+C, \text{where C is some constant.}$