Question

evaluate the following integral.
int\frac{x^{2}+10x - 1}{x^{3}-x}dx

find the partial - fraction decomposition of the integrand.
int\frac{x^{2}+10x - 1}{x^{3}-x}dx=int(square)dx

Explanation:

Step1: Factor the denominator

First, factor $x^{3}-x=x(x^{2} - 1)=x(x - 1)(x + 1)$.

Step2: Set up partial - fraction decomposition

Let $\frac{x^{2}+10x - 1}{x(x - 1)(x + 1)}=\frac{A}{x}+\frac{B}{x - 1}+\frac{C}{x + 1}$. Then $x^{2}+10x - 1=A(x - 1)(x + 1)+Bx(x + 1)+Cx(x - 1)$.

Step3: Find the values of A, B, and C

• If $x = 0$:

Substitute $x = 0$ into $x^{2}+10x - 1=A(x - 1)(x + 1)+Bx(x + 1)+Cx(x - 1)$. We get $-1=A(-1)(1)$, so $A = 1$.

• If $x=1$:

Substitute $x = 1$ into $x^{2}+10x - 1=A(x - 1)(x + 1)+Bx(x + 1)+Cx(x - 1)$. We have $1 + 10-1=B\times1\times(1 + 1)$, so $10 = 2B$ and $B = 5$.

• If $x=-1$:

Substitute $x=-1$ into $x^{2}+10x - 1=A(x - 1)(x + 1)+Bx(x + 1)+Cx(x - 1)$. We get $1-10 - 1=C\times(-1)\times(-2)$, so $-10 = 2C$ and $C=-5$.

Step4: Write the partial - fraction decomposition

So $\frac{x^{2}+10x - 1}{x(x - 1)(x + 1)}=\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1}$. And $\int\frac{x^{2}+10x - 1}{x^{3}-x}dx=\int(\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1})dx$.

Answer:

$\int(\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1})dx$