Evaluate the following integrals: x^3-3x x^4+2x^2-4 dx

Question

Evaluate the following integrals:

$\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$

✓

Answer

$\text{I}=\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$

$=\int\frac{\text{x}(\text{x}^2-3)}{\text{x}^4+2\text{x}^2-4}\text{ dx}$

Let $\text{x}^2=\text{t},$ or, $2\text{x}\text{ dx}=\text{dt}$

$\text{I}=\frac{1}{2}\int\frac{(\text{t}-3)}{\text{t}^2+2\text{t}-4}\text{ dt}$

$=\frac{1}{4}\int\frac{2\text{t}-6}{\text{t}^2+2\text{t}-4}\text{ dt}$

$=\frac{1}{4}\int\frac{2\text{t}+2-8}{\text{t}^2+2\text{t}-4}\text{ dt}$

$=\frac{1}{4}\int\Big(\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}-\frac{8}{\text{t}^2+2\text{t}-4}\Big)\text{dt}$

$=\frac{1}{4}\Big(\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}-\int\frac{8}{\text{t}^2+2\text{t}-4}\text{ dt}\Big)$

$\Rightarrow\text{I}=\frac{1}{4}(\text{I}_1+\text{I}_2)\ ....(1)$

Now,

$\text{I}_1=\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}$

Let $\text{t}^2+2\text{t}-4=\text{u}$

or, $(2\text{t}+2)\text{dt}=\text{du}$

$\Rightarrow\text{I}_1=\int\frac{1}{\text{u}}\text{ du}=\int|\text{u}|+\text{C}_1$

$\Rightarrow\text{I}_1=\ln|\text{t}^2+2\text{t}-4|+\text{C}_1$

$\therefore\ \text{I}_1=\ln|\text{x}^4+2\text{x}^2-4|+\text{C}_1$

Now,

$\text{I}_2=\int\frac{-8}{(\text{t}+1)^2-5}\text{ dt}$

$\Rightarrow\text{I}_2=\int\frac{8}{\big(\sqrt5\big)^2-(\text{t}+1)^2}\text{ dt}$

$\therefore\ \text{I}_2=\frac{8}{2\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}_2$

So, from (1), we get

$\text{I}=\frac{1}{4}\Big[\ln|\text{x}^4+2\text{x}^2-4|+\frac{4}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|\Big]+\text{C}$

$\therefore\ \text{I}=\frac{1}{4}\ln|\text{x}^4+2\text{x}^2-4|+\frac{1}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}$

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