Evaluate the following integrals. int(dx)/(x^3-x^2)

Question
Applications of integrals
asked 2021-01-06
Evaluate the following integrals.
\(\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{{3}}-{x}^{{2}}}}\)

Answers (1)

2021-01-07
Step 1
Consider the provided integral,
\(\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{{3}}-{x}^{{2}}}}\)
Evaluate the following integrals.
We can write as,
\(\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{{3}}-{x}^{{2}}}}=\int\frac{{1}}{{{x}^{{2}}{\left({x}-{1}\right)}}}{\left.{d}{x}\right.}\)
First, we find the partial fraction.
\(\displaystyle\frac{{1}}{{{x}^{{2}}{\left({x}-{1}\right)}}}=\frac{{a}_{{0}}}{{x}}+\frac{{a}_{{1}}}{{x}^{{2}}}+\frac{{a}_{{3}}}{{{x}-{1}}}\)...(1)
Step 2
Now, multiply the equation by the provided denominator.
\(\displaystyle\frac{{{1}\cdot{x}^{{2}}{\left({x}-{1}\right)}}}{{{x}^{{2}}{\left({x}-{1}\right)}}}=\frac{{{a}_{{0}}{x}^{{2}}{\left({x}-{1}\right)}}}{{x}}+\frac{{{a}_{{1}}{x}^{{2}}{\left({x}-{1}\right)}}}{{x}^{{2}}}+\frac{{{a}_{{2}}{x}^{{2}}{\left({x}-{1}\right)}}}{{x}}-{1}\)
\(\displaystyle{1}={a}_{{0}}{x}{\left({x}-{1}\right)}+{a}_{{1}}{\left({x}-{1}\right)}+{a}_{{2}}{x}^{{2}}\)
Compare the both the sides of the coefficient.
\(\displaystyle{a}_{{1}}=-{1},{a}_{{2}}={1}\)
Using above values find the \(\displaystyle{a}_{{0}}\).
\(\displaystyle{a}_{{0}}=-{1}\).
Step 3
Put the above values in the equation (1).
\(\displaystyle\frac{{1}}{{{x}^{{2}}{\left({x}-{1}\right)}}}=\frac{{-{1}}}{{x}}+\frac{{-{1}}}{{x}^{{2}}}+\frac{{1}}{{{x}-{1}}}\)
\(\displaystyle=-\frac{{1}}{{x}}=\frac{{1}}{{x}^{{2}}}+\frac{{1}}{{{x}-{1}}}\)
So, the integrals becomes.
\(\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{{3}}-{x}^{{2}}}}=\int\frac{{1}}{{{x}^{{2}}{\left({x}-{1}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-\int\frac{{1}}{{x}}{\left.{d}{x}\right.}-\int\frac{{1}}{{x}^{{2}}}{\left.{d}{x}\right.}+\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\ln{{\left|{{x}}\right|}}}-{\left(-\frac{{1}}{{x}}\right)}+{\ln{{\left|{{x}-{1}}\right|}}}\)
\(\displaystyle=-{\ln{{\left|{{x}}\right|}}}+\frac{{1}}{{x}}+{\ln{{\left|{{x}-{1}}\right|}}}+{C}\)
Hence.
0

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