Charles Kingsley

2022-01-07

Evaluate the integral.
$\int \frac{1}{x\left(x-a\right)}dx$

psor32

Step 1
Given data:
The integral expression is: $\int \frac{1}{x\left(x-a\right)}dx$
The given integral can be expressed as,
$\int \frac{1}{x\left(x-a\right)}dx=\int \frac{1}{ax\left(\frac{x}{a}-1\right)}dx$
$=\frac{1}{a}\int \frac{1}{x\left(\frac{x}{a}-1\right)}dx$
Assume x/a =u, and differentiate it with respect to x.
$\frac{1}{a}dx=du$
Step 2
Substitute the above-calculated values in the expression.
$\int \frac{1}{x\left(x-a\right)}dx=\frac{1}{a}\int \frac{1}{\left(au\right)\left(u-1\right)}\left(adu\right)$
$=\frac{1}{a}\int \frac{du}{u\left(u-1\right)}$
$=\frac{1}{a}\int \left(\frac{-1}{u}+\frac{1}{u-1}\right)du$
$=\frac{1}{a}\left(-\mathrm{ln}|u|+\mathrm{ln}|u-1|\right)+C$
$=\frac{1}{a}\mathrm{ln}|\frac{u-1}{u}|+C$
Substitute x/a for u in the expression.
$\int \frac{1}{x\left(x-a\right)}dx=\frac{1}{a}\mathrm{ln}|\frac{\left(\frac{x}{a}\right)-1}{\left(\frac{x}{a}\right)}|+C$
$=\frac{1}{a}\mathrm{ln}|\frac{x-a}{x}|+C$
Thus, the integration of the given expression is $\frac{\mathrm{ln}|\frac{x-a}{x}|}{a}+C$.

Dawn Neal

Step 1
We can write
$\frac{1}{x\left(x-a\right)}=\frac{A}{x}+\frac{B}{x-a}$
Multiply both sides by x(x-a)
1=A(x-a)+Bx
1=Ax-Aa+Bx
Group the like terms
0*x+1=(A+B)x-Aa
1=-Aa
Divide both sides by -a
$-\frac{1}{a}=A$
0=A+B
Substitute A with $-\frac{1}{a}$, to get
$0=-\frac{1}{a}+B$
Add $\frac{1}{a}$ to both sides
$\frac{1}{a}=B$
Step 2
Substitute back the values of A and B, to get
$\int \frac{1}{x\left(x-a\right)}dx=\int \frac{\frac{1}{a}}{x}-\frac{\frac{1}{a}}{x-a}dx$
$=\frac{1}{a}\mathrm{ln}|x|-\frac{1}{a}\mathrm{ln}|x-a|+C$
$=\frac{1}{a}\mathrm{ln}|\frac{x}{x-a}|+C$
Result:
$\int \frac{1}{x\left(x-a\right)}dx=\frac{1}{a}\mathrm{ln}|\frac{x}{x-a}|+C$

karton