Answer to Question #223110 in Calculus for Varylan

Question

Evaluate the following Integrals

1)  ʃ [2x² / (x-1)²(x+1)] dx

2)  ʃ [2x³-3x²-x-7 / 2x²-3x-2] dx

Answer 1

1) Let us expand the integrand into partial fractions:

$\frac{{2{x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x + 1}} = \frac{{A({x^2} - 1) + B(x + 1) + C\left( {{x^2} - 2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \frac{{{x^2}\left( {A + C} \right) + x\left( {B - 2C} \right) - A + B + C}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}$

$\left\{ \begin{array}{l} A + C = 2\\ B - 2C = 0\\ - A + B + C = 0 \end{array} \right. \Rightarrow A = \frac{3}{2},\,\,B = 1,\,\,C = \frac{1}{2}$

Then

$\int {\frac{{2{x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}dx} = \frac{3}{2}\int {\frac{{dx}}{{x - 1}} + \int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} } + \frac{1}{2}\int {\frac{{dx}}{{x + 1}}} = \frac{3}{2}\ln |x - 1| - \frac{1}{{x - 1}}\_\frac{1}{2}\ln |x + 1| + C$

Answer: $\frac{3}{2}\ln |x - 1| - \frac{1}{{x - 1}}\_\frac{1}{2}\ln |x + 1| + C$

2)

$\int {\frac{{2{x^3} - 3{x^2} - x - 7}}{{2{x^2} - 3x - 2}}dx = } \int {\frac{{2{x^3} - 3{x^2} - 2x + x - 7}}{{2{x^2} - 3x - 2}}dx = } \int {\frac{{2{x^3} - 3{x^2} - 2x}}{{2{x^2} - 3x - 2}}dx + \int {\frac{{x - 7}}{{2{x^2} - 3x - 2}}} } dx = \int {\frac{{x\left( {2{x^2} - 3x - 2} \right)}}{{2{x^2} - 3x - 2}}dx + \frac{1}{4}\int {\frac{{4x - 28}}{{2{x^2} - 3x - 2}}} } dx = \int {xdx + \frac{1}{4}\int {\frac{{4x - 3 - 25}}{{2{x^2} - 3x - 2}}} dx} = \int {xdx + \frac{1}{4}\int {\frac{{4x - 3}}{{2{x^2} - 3x - 2}}} dx} - \frac{{25}}{4}\int {\frac{{dx}}{{2{x^2} - 3x - 2}}} = \int {xdx + \frac{1}{4}\int {\frac{{d\left( {2{x^2} - 3x - 2} \right)}}{{2{x^2} - 3x - 2}}} } - \frac{{25}}{8}\int {\frac{{dx}}{{{x^2} - \frac{3}{2}x - 1}}} = \int {xdx + \frac{1}{4}\int {\frac{{d\left( {2{x^2} - 3x - 2} \right)}}{{2{x^2} - 3x - 2}}} } - \frac{{25}}{8}\int {\frac{{dx}}{{{x^2} - \frac{3}{2}x + \frac{9}{{16}} - \frac{{25}}{{16}}}} = } \int {xdx + \frac{1}{4}\int {\frac{{d\left( {2{x^2} - 3x - 2} \right)}}{{2{x^2} - 3x - 2}}} } + \frac{{25}}{8}\int {\frac{{dx}}{{{{\left( {\frac{5}{4}} \right)}^2} - {{\left( {x - \frac{3}{4}} \right)}^2}}} = } \frac{{{x^2}}}{2} + \frac{1}{4}\ln \left| {2{x^2} - 3x - 2} \right| + \frac{{25}}{8} \cdot \frac{1}{{2 \cdot \frac{5}{4}}}\ln \left| {\frac{{\frac{5}{4} + x - \frac{3}{4}}}{{\frac{5}{4} - \left( {x - \frac{3}{4}} \right)}}} \right| + C = \frac{{{x^2}}}{2} + \frac{1}{4}\ln \left| {2{x^2} - 3x - 2} \right| + \frac{5}{4}\ln \left| {\frac{{\frac{1}{2} + x}}{{2 - x}}} \right| + C$

Answer" $\frac{{{x^2}}}{2} + \frac{1}{4}\ln \left| {2{x^2} - 3x - 2} \right| + \frac{5}{4}\ln \left| {\frac{{\frac{1}{2} + x}}{{2 - x}}} \right| + C$