Question

Evaluate the integral \(\int {\frac{{dx}}{{2{x^3} - 3{x^2}}}} \)

Short answer

\(\)\(\frac{1}{{3x}} - \frac{{2\ln (x)}}{9} + \frac{{2\ln (2x - 3)}}{9}\)is the value of given integral.

Step-by-step solution

Step 1-Simply the fraction

Using Partial fraction decomposition method to solve this question.

\(\begin{aligned}{l}\int {\frac{{dx}}{{2{x^3} - 3{x^2}}} = \int {\frac{1}{{{x^2}(2x - 3)}}} } .dx\\\end{aligned}\)

\(\int {(\frac{4}{{9(2x - 3)}} - \frac{2}{{9x}}} - \frac{1}{{3{x^2}}}).dx\)

\(\frac{4}{9}\int {\frac{{dx}}{{2x - 3}} - \frac{2}{9}} \int {\frac{{dx}}{x}} - \frac{1}{3}\int {\frac{{dx}}{{{x^2}}}} \)

– To Find the integral

= \(\begin{aligned}{l}\frac{4}{9}\int {\frac{{dx}}{{2x - 3}}} - \frac{2}{9}\int {\frac{{dx}}{x} - \frac{1}{3}\int {\frac{{dx}}{{{x^2}}}} } \\\end{aligned}\)

=Let\(t = 2x - 3\)

Differentiating both sides with respect to x

\(dt = 2dx\),\(dx = \frac{{dt}}{2}\)

=\(\frac{4}{9}\int {\frac{1}{t}} *\frac{{dt}}{2} - \frac{2}{9}(\ln (x)) - \frac{1}{3}( - \frac{1}{x})\)

=\(\frac{2}{9}\ln t - \frac{{2\ln (x)}}{9} + \frac{1}{{3x}}\)

=\(\frac{{2\ln (2x - 3)}}{9} - \frac{{2\ln (x)}}{9} + \frac{1}{{3x}}\)

=\(\frac{1}{{3x}} - \frac{{2\ln (x)}}{9} + \frac{{2\ln (2x - 3)}}{9}\)

Hence,The Value of Given integral is\(\frac{1}{{3x}} - \frac{{2\ln (x)}}{9} + \frac{{2\ln (2x - 3)}}{9}\)..