Find the integral int((x^2)/(x^4-2x^2+8))dx

 Exercise

$\int\frac{x^{2}}{x^{4}-2x^{2}+8}dx$

 

We can factor the fourth degree trinomial $x^4-2x^2+8$ by applying the substitution: $y=x^2$

$\int\frac{x^2}{y^2-2y+8}dx$

 

Take the constant $\frac{1}{y^2-2y+8}$ out of the integral

$\frac{1}{y^2-2y+8}\int x^2dx$

 

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$\frac{1}{y^2-2y+8}\frac{x^{3}}{3}$

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Multiplying fractions $\frac{1}{y^2-2y+8} \times \frac{x^{3}}{3}$

$\frac{x^{3}}{3\left(y^2-2y+8\right)}$

 

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{x^{3}}{3\left(y^2-2y+8\right)}+C_0$

$\frac{x^{3}}{3\left(y^2-2y+8\right)}+C_0$

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