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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals3 Marks
Question
Evaluate the following definite integrals:$\int_{0}^\limits{1}\frac{1}{2\text{x}^2+\text{x}+1}\text{ dx}$
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Answer

We have,$\int_{0}^\limits{1}\frac{1}{2\text{x}^2+\text{x}+1}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{1\text{ dx}}{\big(\text{x}^2+\frac{1}{2}\text{x}+\frac{1}{2}\big)}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\frac{1}{2}-\frac{1}{16}}$ $\Big[\text{Adding }\frac{1}{16}\&\text{ substracting }\frac{1}{16}\text{ in numerator}\Big]$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\frac{7}{16}}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\big(\frac{\sqrt{7}}{4}\big)^2}$
$=\frac{1}{2}\cdot\frac{4}{\sqrt{7}}\Bigg[\tan^{-1}\Bigg(\frac{\text{x}+\frac{1}{4}}{\frac{\sqrt{7}}4{}}\Bigg)\Bigg]^1_0$
$=\frac{2}{\sqrt{7}}\bigg\{\tan^{-1}\frac{5}{\sqrt{7}}-\tan^{-1}\Big(\frac{1}{\sqrt{7}}\Big)\bigg\}$

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