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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$
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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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