Question

Evaluate the following integral:

\[\int\frac{1}{x\left( x^3 + 8 \right)}dx\]

 

Answer 1

\[\text{Let }I = \int\frac{1}{x\left( x^3 + 8 \right)}dx\]
We express
\[\frac{1}{x\left( x^3 + 8 \right)} = = \frac{A}{x} + \frac{B x^2 + Cx + D}{x^3 + 8}\]
\[ \Rightarrow 1 = A\left( x^3 + 8 \right) + \left( B x^2 + Cx + D \right)\left( x \right)\]
Equating the coefficients of `x^3 , x^2 , x` and constants, we get

\[0 = A + B\text{ and }0 = C\text{ and }0 = D\text{ and }1 = 8A\]
\[\text{or }A = \frac{1}{8}\text{ and }B = - \frac{1}{8}\text{ and }C = 0\text{ and }D = 0\]
\[ \therefore I = \int\left( \frac{\frac{1}{8}}{x} + \frac{- \frac{1}{8} x^2}{\left( x^3 + 8 \right)} \right)dx\]
\[ = \frac{1}{8}\int\frac{1}{x}dx - \frac{1}{24}\int\frac{3 x^2}{x^3 + 8} dx\]
\[ = \frac{1}{8}\log\left| x \right| - \frac{1}{24}\log\left| x^3 + 8 \right| + c\]
\[\text{Hence, }\int\frac{1}{x\left( x^3 + 8 \right)}dx = \frac{1}{8}\log\left| x \right| - \frac{1}{24}\log\left| x^3 + 8 \right| + c\]