Evaluate the following integral:
Question:

Evaluate the following integral:

$\int \frac{1}{x\left(x^{3}+8\right)} d x$

Solution:

Consider the integral,

$I=\int \frac{1}{x\left(x^{3}+8\right)} d x$

Rewriting the above integral, we have

$I=\int \frac{x^{2}}{x^{3}\left(x^{3}+8\right)} d x$

$I=\frac{1}{3} \int \frac{3 x^{2}}{x^{3}\left(x^{3}+8\right)} d x$

Substitute $x^{3}=t$

$3 x^{2} d x=d t$

$I=\frac{1}{3} \int \frac{d t}{t(t+8)}$

$\frac{1}{t(t+8)}=\frac{A}{t}+\frac{B}{t+8}$

$1=A(t+8)+B t$

Equating constants

$1=8 \mathrm{~A}$

$\mathrm{~A}=\frac{1}{8}$

Equating coefficients of $t$

$0=A+B$

$B=-\frac{1}{8}$

$I=\frac{1}{3} \int \frac{d t}{t(t+8)}$

$=\frac{1}{3} \int\left(\frac{\frac{1}{8}}{t}-\frac{\frac{1}{8}}{t+8}\right) d t$

$=\frac{1}{3} \times \frac{1}{8} \int \frac{\mathrm{dt}}{\mathrm{t}}-\frac{1}{3} \times \frac{1}{8} \int \frac{\mathrm{dt}}{\mathrm{t}+8}$

$=\frac{1}{24} \log \mathrm{t}-\frac{1}{24} \log |\mathrm{t}+8|+\mathrm{C}$

$=\frac{1}{24} \log \mathrm{x}^{3}-\frac{1}{24} \log \left|\mathrm{x}^{3}+8\right|+\mathrm{C}$

$=\frac{3}{24} \log \mathrm{x}-\frac{1}{24} \log \left|\mathrm{x}^{3}+8\right|+\mathrm{C}$

$=\frac{1}{8} \log \mathrm{x}-\frac{1}{24} \log \left|\mathrm{x}^{3}+8\right|+\mathrm{C}$

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