Question

Evaluate the integral:\(\int_0^1 {\frac{2}{{2{x^2} + 3x + 1}}} dx\)

Short answer

The value of the integral is \(\int\limits_0^1 {\frac{{2\,dx}}{{2{x^2} + 3x + 1}} = 2\,log\left( {\frac{3}{2}} \right)} \)\(\)

Step-by-step solution

Factorizing the denominator.

\(2{x^2} + 3x + 1 = 2{x^2} + 2x + 1x + 1\)

\( = 2x(x + 1) + 1(x + 1)\)

\( = (2x + 1)\left( {x + 1} \right)\)

Finding the Integral

\(\int_0^1 {\frac{{2dx}}{{2{x^2} + 3x + 1}}} = \int_0^1 {\frac{{2dx}}{{(2x + 1)\left( {x + 1} \right)}}} \)

Now, using partial fraction,

\(\frac{2}{{(2x + 1)\left( {x + 1} \right)}} = \frac{A}{{2x + 1}} + \frac{B}{{x + 1}}.\)

\(\frac{2}{{\left( {2x + 1} \right)\left( {x + 1} \right)}} = \frac{{A(x + 1) + B(2x + 1)}}{{\left( {2x + 1} \right)\left( {x + 1} \right)}}\)

Comparing both sides.

\(2 = A(x + 1) + B(2x + 1)\)

\(2 = Ax + A + 2Bx + B\)

\(2 = (A + 2B)x + A + B\)

\( \Rightarrow A + 2B = 0\) and \(A + B = 2\).

Solving the two equations.

\(A + 2B = 0\)&\(A + B = 2\)

\(\begin{array}{l}A + 2B = 0\\A + B = 2\\ - - - \\\_\_\_\_\_\_\_\_\_\\2B - B = 0 - 2\\B = - 2\end{array}\)

And\(A = - 2B = + 4\)

\(\frac{A}{{2x + 1}} + \frac{B}{{x + 1}} = \frac{4}{{2x + 1}} + \frac{{( - 2)}}{{x + 1}}\)

\(\int {\frac{{4dx}}{{2x + 1}}} - \int {\frac{{2dx}}{{x + 1}}} \)

Solving them separately.

\(\int {\frac{{4dx}}{{2x + 1}}} = 4 \int {\frac{{dx}}{{2x + 1}}} \)

Let

\(2x + 1 = t\)

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

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

\( \Rightarrow 4\int {\frac{1}{t}} \times \frac{{dt}}{2} = \frac{4}{2}log|t|\)

\( = 2log|t| + {c_1}\)

\( = 2log|2x + 1| + {c_1}\)

\(\int {\frac{{2dx}}{{x + 1}}} = 2\int {\frac{{dx}}{{x + 1}}} \)

Let\( x + 1 = u\)

\(1 = \frac{{du}}{{dx}}\)

\( \Rightarrow du = dx\)

\(2\int {\frac{{du}}{u}} = 2log|u| + {c_2}\)

\( = 2log|x + 1| + {c_2}\)

\(\int {\frac{{4dx}}{{2x + 1}}} - \int {\frac{{2dx}}{{x + 1}}} = 2log|2x + 1| + {c_1} - 2log|x + 1| + {c_2}\)

\( = 2(log|2x + 1| - log|x + 1|) + {c_1} + {c_2}\)

\( = 2\left( {log\left( {\frac{{2x + 1}}{{x + 1}}} \right)} \right) + {c_1} + {c_2}\)

\(\left( { as logm - logn = log\left( {\frac{m}{n}} \right)} \right)\)

\( = 2\left( {log\left( {\frac{{2x + 1}}{{x + 1}}} \right)} \right) + c\)

(as \({c_1} + {c_2}\) is another constant i.e c)

Putting the limits.

\( = \left( {2log\left( {\frac{{2x + 1}}{{x + 1}}} \right)} \right)_0^1\)

\( = 2log\left( {\frac{{2 + 1}}{{1 + 1}}} \right) - 2log\left( {\frac{1}{1}} \right)\)

\( = 2log\left( {\frac{3}{2}} \right) - 2log\left( 1 \right)\)

\( = 2\,\log \left( {\frac{3}{2}} \right)\) (as\(log\left( 1 \right) = 0\))

Hence, the value of the integral is \(\int\limits_0^1 {\frac{{2\,dx}}{{2{x^2} + 3x + 1}} = 2\,log\left( {\frac{3}{2}} \right)} \)\(\)