Question
Evaluate the following integrals:
$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
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Answer
$\text{I}=\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$ $=\int\frac{\text{x}^2\cdot\text{x}}{(\text{x}^2)^2+\text{x}^2+1}\text{ dx}$ Let $\text{x}^2=\text{t}$ or $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\frac{2\text{t}+1-1}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\Big[\frac{(2\text{t}+1)}{(\text{t}^2+\text{t}+1)}-\frac{1}{(\text{t}^2+\text{t}+1)}\Big]\text{dt}$
$=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}\big)}\text{ dt}\Big]$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{ dt}\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\frac{\big(\text{t}+\frac{1}{2}\big)}{\Big(\frac{\sqrt3}{2}\Big)}\end{bmatrix}+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{t}+1}{\sqrt3}\Big)\Big]+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{x}^4+\text{x}^2+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{x}^2+1}{\sqrt3}\Big)\Big]+\text{C}$Need a full question paper?
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