Integral Calculus
86291
\(\int \frac{d x}{x+\sqrt{x-1}}=\)
1 \(\log _{\mathrm{e}}|\mathrm{x}+\sqrt{\mathrm{x}-1}|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{\mathrm{x}-1}+1}{\sqrt{3}}\right)+\mathrm{c}\)
2 \(\frac{1}{\sqrt{3}} \log _{e}|x+\sqrt{x-1}|-\tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
3 \(\frac{2}{\sqrt{3}} \log _{\mathrm{e}}|\mathrm{x}+\sqrt{\mathrm{x}-1}|-\tan ^{-1}\left(\frac{2 \sqrt{\mathrm{x}-1}+1}{\sqrt{3}}\right)+\mathrm{c}\)
4 \(\log _{\mathrm{e}}|\mathrm{x}+\sqrt{\mathrm{x}-1}|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{\mathrm{x}-1}+1}{\sqrt{3}}\right)+\mathrm{c}\)
Explanation:
(D) : Given,
\(I=\int \frac{d x}{x+\sqrt{x-1}}\)
Let,
\(\mathrm{x}-1=\mathrm{t}^{2}\)
\(\mathrm{x}=\mathrm{t}^{2}+1\)
\(\mathrm{dx}=2 \mathrm{tdt}\)
So, \(\quad I=2 \int \frac{t d t}{t^{2}+1+t}=\int \frac{2 t+1-1}{t^{2}+t+1} d t\)
\(=\int \frac{(2 t+1) d t}{t^{2}+t+1}-1 \int \frac{d t}{t^{2}+t+1}\)
Let, \(\quad I_{1}=\int \frac{2 t+1}{t^{2}+t+1} d t\)
\(\mathrm{I}_{1}=\log \left(\mathrm{t}^{2}+\mathrm{t}+1\right)\)
\(I_{2}=\int \frac{1}{t^{2}+t+\frac{1}{4}+1-\frac{1}{4}} d t\)
\(I_{2}=\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d t\)
\(\because \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+\mathrm{a}^{2}}=\frac{1}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}\)
\(\therefore \mathrm{I}_{2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\mathrm{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\mathrm{C}\)
\(\mathrm{I}=\mathrm{I}_{1}-\mathrm{I}_{2}\)
\(I=\log \left(t^{2}+t+1\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C\)
\(=\log \left(t^{2}+t+1\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+C\)
On putting value of \(t=\sqrt{x-1}\)
\(I=\log |(x+\sqrt{x-1})|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C\)
