TS EAMCET · Maths · Indefinite Integration
\(\int \sqrt{x^2+x+1} d x \times \int \frac{1}{\sqrt{x^2+x+1}} d x\) is equal to
- A x + C
- B \(\begin{aligned}\left(\frac{2 x+1}{4} \sqrt{x^2+x+1}+\frac{3}{8} \sinh ^{-1} \frac{2 x+1}{\sqrt{3}}\right) & \ & \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C\end{aligned}\)
- C \(\begin{array}{r}\frac{2 x+1}{2} \sinh ^{-1}\left(\sqrt{x^2+x+1}\right)+\left(\frac{3}{8} \sinh ^{-1} \frac{2 x+1}{\sqrt{3}}\right)^2 \ +C\end{array}\)
- D \(\frac{2 x+1}{2}\left(\sinh ^{-1} \frac{2 x+1}{\sqrt{3}}\right)^2+C\)
Answer & Solution
Correct Answer
(B) \(\begin{aligned}\left(\frac{2 x+1}{4} \sqrt{x^2+x+1}+\frac{3}{8} \sinh ^{-1} \frac{2 x+1}{\sqrt{3}}\right) & \ & \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C\end{aligned}\)
Step-by-step Solution
Detailed explanation
\(\because x^2+x+1=\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\)…
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