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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:
$\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
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Answer

Let $\text{I}=\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
$=\int\sqrt{\text{x}^2+\text{x}+\frac{1}{4}+\frac{3}{4}}\text{dx}$
$=\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{dx}$
$=\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}+\frac{\big(\frac{\sqrt{3}}{2}\big)^2}{2}\times\\\log\Bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\Bigg|+\text{C}$
$=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|\Big(\frac{2\text{x}+1}{2}\Big)+\frac{1}{2}\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
$\therefore\ \text{I}=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|2\text{x}+1+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$

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