Question

Find the integral value of \(I = \int {\frac{1}{{x{{\left( {{x^2} + 4} \right)}^2}}}} \)

Short answer

Multiply and divide I by x and thus simplify using partial fractions.

Step-by-step solution

Step 1: Substitution and reducing

\(I = \int {\frac{1}{{x{{\left( {{x^2} + 4} \right)}^2}}}} .\frac{x}{x}\)

\( = \int {\frac{x}{{{x^2}\left( {{x^2} + 4} \right)}}dx} \)

\( = \frac{1}{2}\int {\frac{{2xdx}}{{{x^2}{{\left( {{x^2} + 4} \right)}^2}}}} \)

Put \({x^2} = t\) and \(2xdx = dt\)

\(I = \frac{1}{2}\int {\frac{{dt}}{{t{{(t + 4)}^2}}}} \)

Partial fraction simplification

\(\frac{{dt}}{{t{{\left( {t + 4} \right)}^2}}} = \frac{A}{t} + \frac{B}{{t + 4}} + \frac{C}{{{{\left( {t + 4} \right)}^2}}}\)

\(1 = A{\left( {t + 4} \right)^2} + Bt\left( {t + 4} \right) + Ct\)

For For

For \(\begin{array}{l}t = 1\\1 = 25A + 5B + C\\1 = \frac{{25}}{{16}} + 5B - \frac{1}{4}\\1 = \frac{{21}}{{16}} + 5B \Rightarrow B = \frac{{ - 1}}{{16}}\end{array}\)

Substitution the values

\(\therefore I = \int {\frac{{dt}}{{16t}} - \int {\frac{{dt}}{{16\left( {t + 4} \right)}} - \int {\frac{{dt}}{{4{{\left( {t + 4} \right)}^2}}}} } } \)

Simplification

\(I = \int {\frac{{dt}}{{16t}} + \int {\frac{{ - dt}}{{16\left( {t + 4} \right)}} + \int {\frac{{ - dt}}{{4{{\left( {t + 4} \right)}^2}}}} } } \)

\( = \frac{1}{{16}}\log t - \frac{1}{{16}}\log |t + 4| - \frac{1}{4}.\frac{1}{{t + 4}}\)

\( = \frac{1}{{16}}\log |{x^2}| - \frac{1}{{16}}\log |{x^2} + 4| - \frac{1}{{4\left( {{x^2} + 4} \right)}}\)

Hence, \(I = \int {\frac{{dx}}{{x{{\left( {{x^2} + 4} \right)}^2}}} = \frac{1}{{16}}\log \left( {\frac{{{x^2}}}{{{x^{2 + }}4}}} \right) - \frac{1}{{4\left( {{x^2} + 4} \right)}}} \)