Question

Evaluate the integral\(\int {\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + x}}}}} \)

Short answer

Decompose partial fractions with partial fraction decomposition \(\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + x}}}}{\rm{ = ln|x| - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ln}}\left| {{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}} \right|{\rm{ + C}}\).

Step-by-step solution

Partial fraction decomposition is a method for separating partial fractions.

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

Partial fraction\({{\rm{x}}^{\rm{2}}}{\rm{ + 1}}\)decomposition is possible because is an irreducible quadratic polynomial with no real roots.

\(\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + x}}}}{\rm{ = }}\frac{{\rm{A}}}{{\rm{x}}}{\rm{ + }}\frac{{{\rm{Bx + C}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}}}{\rm{.}}\)

Divide by the smallest common denominator\({{\rm{x}}^{\rm{3}}}{\rm{ + x}}\), we obtain

\(\begin{aligned}{c}{\rm{1 = A}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}} \right){\rm{ + (Bx + C)x}}\\{\rm{1 = A}}{{\rm{x}}^{\rm{2}}}{\rm{ + A + B}}{{\rm{x}}^{\rm{2}}}{\rm{ + Cx}}\\{\rm{1 = (A + B)}}{{\rm{x}}^{\rm{2}}}{\rm{ + Cx + A}}\end{aligned}\)

Left and right side polynomials.

The coefficients of the polynomials on the left and right sides are now equated.

\(\begin{aligned}{c}{\rm{A = 1}}\\{\rm{C = 0 }}\\{\rm{A + B = 0}}\\{\rm{B = - A = - 1}}\end{aligned}\)

\(\begin{aligned}{c}\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + x}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - }}\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}}}\;\;\\\;{\rm{dx = }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{dx - }}\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}}}{\rm{dx}}\\{\rm{ = ln|x| - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ln}}\left| {{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}} \right|{\rm{ + C}}{\rm{.}}\end{aligned}\)

We are now going to calculate the supplied integral\(\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + x}}}}{\rm{ = ln|x| - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ln}}\left| {{{\rm{x}}^{\rm{2}}}{\rm{ + 1}}} \right|{\rm{ + C}}{\rm{.}}\)