Correct Answer is : log⁡e∣x+x−1∣_{e} |x+{x-1}|loge∣x+x−1∣ - 23tan⁡−1(2x−1+13){2}{{3}} ^{-1}≤ft({2{x-1}+1}{{3}})32tan−1(32x−1+1) + c

_{e} |x+{x-1}| - {1}{{3}} ^{-1}≤ft({2{x-1}+1}{{3}}) + c

{1}{{3}} _{e} |x+{x-1}| - ^{-1}≤ft({2{x-1}+1}{{3}}) + c

{2}{{3}} _{e} |x+{x-1}| - ^{-1}≤ft({2{x-1}+1}{{3}}) + c

_{e} |x+{x-1}| - {2}{{3}} ^{-1}≤ft({2{x-1}+1}{{3}}) + c

Correct Answer is : log⁡e∣x+x−1∣_{e} |x+{x-1}|loge∣x+x−1∣ - 23tan⁡−1(2x−1+13){2}{{3}} ^{-1}≤ft({2{x-1}+1}{{3}})32tan−1(32x−1+1) + c

Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

∫

\frac{dx}{x+\sqrt{x-1}}

$\log_{e} |x+\sqrt{x-1}|$ - $\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right)$ + c
$\frac{1}{\sqrt{3}} \log_{e} |x+\sqrt{x-1}|$ - $\tan^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right)$ + c
$\frac{2}{\sqrt{3}} \log_{e} |x+\sqrt{x-1}|$ - $\tan^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right)$ + c
$\log_{e} |x+\sqrt{x-1}|$ - $\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right)$ + c

Solution:

Consider the integral

I=\int\;\frac{dx}{x+\sqrt{x-1}}

Substitute these values in the integral and solve

I=\int\;\frac{2t}{(t^{2}+1)+t} dt

=\int\;\frac{(2t+1)-1}{t^{2}+t+1} dt

=\int\;\frac{2t+1}{t^{2}+t+1} dt - \int\left(\;\frac{dt}{t^{2}+t+1}\right)

=\log_{e}|t^{2}+t+1| - \int\;\frac{dt}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}

Further simplify the above

I=\log_{e}|t^{2}+t+1| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + c

\;\;\;\; = \log_{e}|t^{2}+t+1| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t+1}{\sqrt{3}}\right) + c

Substitute the value of

t=\sqrt{x-1}

I=\log_{e}|x+\sqrt{x-1}| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right) + c

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