AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x+1}{\sqrt{x^2+x+1}} d x=\)
- A \(\frac{1}{2} \sqrt{x^2+x+1}+\frac{1}{2} \cosh ^{-1}\left(\frac{x+2}{\sqrt{3}}\right)+c\)
- B \(\frac{1}{2} \sqrt{x^2+x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
- C \(\sqrt{x^2+x+1}+\frac{2}{\sqrt{3}} \log \left|x^2+x+1\right|+c\)
- D \(\sqrt{x^2+x+1}+\frac{1}{2} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
Answer & Solution
Correct Answer
(D) \(\sqrt{x^2+x+1}+\frac{1}{2} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c\)
Step-by-step Solution
Detailed explanation
Let, \(\mathrm{I}=\int \frac{x+1}{\sqrt{x^2+x+1}} d x\) \(\begin{aligned} & =\int \frac{2 x+1+1}{2 \sqrt{x^2+x+1}} d x \\ & =\int \frac{2 x+1}{2 \sqrt{x^2+x+1}} d x+\frac{1}{2} \int \frac{1}{\sqrt{x^2+x+1}} d x\end{aligned}\) \(=\mathrm{I}_1+\mathrm{I}_2\) Now,…
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