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

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

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psor32
Step 1
Given data:
The integral expression is: $$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}$$
The given integral can be expressed as,
$$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{a}{x}{\left({\frac{{{x}}}{{{a}}}}-{1}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}\int{\frac{{{1}}}{{{x}{\left({\frac{{{x}}}{{{a}}}}-{1}\right)}}}}{\left.{d}{x}\right.}$$
Assume x/a =u, and differentiate it with respect to x.
$$\displaystyle{\frac{{{1}}}{{{a}}}}{\left.{d}{x}\right.}={d}{u}$$
Step 2
Substitute the above-calculated values in the expression.
$$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{a}}}}\int{\frac{{{1}}}{{{\left({a}{u}\right)}{\left({u}-{1}\right)}}}}{\left({a}{d}{u}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}\int{\frac{{{d}{u}}}{{{u}{\left({u}-{1}\right)}}}}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}\int{\left({\frac{{-{1}}}{{{u}}}}+{\frac{{{1}}}{{{u}-{1}}}}\right)}{d}{u}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}{\left(-{\ln}{\left|{u}\right|}+{\ln}{\left|{u}-{1}\right|}\right)}+{C}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}{\ln}{\left|{\frac{{{u}-{1}}}{{{u}}}}\right|}+{C}$$
Substitute x/a for u in the expression.
$$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{a}}}}{\ln}{\left|{\frac{{{\left(\frac{{x}}{{a}}\right)}-{1}}}{{{\left(\frac{{x}}{{a}}\right)}}}}\right|}+{C}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}{\ln}{\left|{\frac{{{x}-{a}}}{{{x}}}}\right|}+{C}$$
Thus, the integration of the given expression is $$\displaystyle\frac{{{\ln}{\left|\frac{{{x}-{a}}}{{x}}\right|}}}{{a}}+{C}$$.
###### Not exactly what youâ€™re looking for?
Dawn Neal
Step 1
We can write
$$\displaystyle{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}={\frac{{{A}}}{{{x}}}}+{\frac{{{B}}}{{{x}-{a}}}}$$
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
$$\displaystyle-{\frac{{{1}}}{{{a}}}}={A}$$
0=A+B
Substitute A with $$\displaystyle-{\frac{{{1}}}{{{a}}}}$$, to get
$$\displaystyle{0}=-{\frac{{{1}}}{{{a}}}}+{B}$$
Add $$\displaystyle{\frac{{{1}}}{{{a}}}}$$ to both sides
$$\displaystyle{\frac{{{1}}}{{{a}}}}={B}$$
Step 2
Substitute back the values of A and B, to get
$$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}=\int{\frac{{\frac{{1}}{{a}}}}{{{x}}}}-{\frac{{\frac{{1}}{{a}}}}{{{x}-{a}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}{\ln}{\left|{x}\right|}-{\frac{{{1}}}{{{a}}}}{\ln}{\left|{x}-{a}\right|}+{C}$$
$$\displaystyle={\frac{{{1}}}{{{a}}}}{\ln}{\left|{\frac{{{x}}}{{{x}-{a}}}}\right|}+{C}$$
Result:
$$\displaystyle\int{\frac{{{1}}}{{{x}{\left({x}-{a}\right)}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{a}}}}{\ln}{\left|{\frac{{{x}}}{{{x}-{a}}}}\right|}+{C}$$
karton

$$\text{Given:} \\\int \frac{1}{x(x-a)}dx \\=\int \frac{1}{(1-\frac{a}{x})}x^{2}dx \\=\frac{1}{a}\int \frac{1}{u}du \\\int \frac{\ln(u)}{a} \\=\frac{\ln(1-\frac{a}{x})}{a} \\=\frac{\ln(|\frac{a}{x}-1|)}{a}+C \\\text{simplify:} \\\frac{\ln(|x-a|)-\ln(|x|)}{a}+C\ \text{(Answer)}$$