Q117 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { The value of } \int \dfrac{1}{x+\sqrt{x-1}} d x \text { is }$
- A$\log (x+\sqrt{x-1})+C$
- B$\log (x+\sqrt{x-1})+\sin ^{-1}\left(\sqrt{\dfrac{x-1}{x}}\right)+C$
- C$\log (x-1+\sqrt{x-1})+\dfrac{1}{\sqrt{3}} \log \left|\dfrac{x-2-\sqrt{3}}{x-2+\sqrt{3}}\right|+C$
- Correct answerD$\log (x+\sqrt{x-1})-\dfrac{2}{\sqrt{3}} \tan ^{-1}\left(\dfrac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$
Solution
Let $I = \int \dfrac{1}{x + \sqrt{x-1}} dx$.
Substitute $t^2 = x - 1$, so $x = t^2 + 1$ and $dx = 2t dt$.
The integral becomes $I = \int \dfrac{2t}{t^2 + 1 + t} dt = \int \dfrac{2t}{t^2 + t + 1} dt$.
Rewrite the numerator as $2t + 1 - 1$:
$I = \int \dfrac{2t + 1}{t^2 + t + 1} dt - \int \dfrac{1}{t^2 + t + 1} dt$.
The first part is $\ln(t^2 + t + 1)$.
For the second part, complete the square in the denominator: $t^2 + t + 1 = (t + \dfrac{1}{2})^2 + \dfrac{3}{4}$.
Using $\int \dfrac{1}{u^2 + a^2} du = \dfrac{1}{a} \tan^{-1}(\dfrac{u}{a})$, we get $\int \dfrac{1}{(t + 1/2)^2 + (\sqrt{3}/2)^2} dt = \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2t + 1}{\sqrt{3}})$.
Thus, $I = \ln(t^2 + t + 1) - \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2t + 1}{\sqrt{3}}) + C$.
Substituting $t = \sqrt{x-1}$ back, $t^2 + t + 1 = x - 1 + \sqrt{x-1} + 1 = x + \sqrt{x-1}$.
Therefore, $I = \ln(x + \sqrt{x-1}) - \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2\sqrt{x-1} + 1}{\sqrt{3}}) + C$.
Answer: $\log (x+\sqrt{x-1})-\dfrac{2}{\sqrt{3}} \tan ^{-1}\left(\dfrac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$
Substitute $t^2 = x - 1$, so $x = t^2 + 1$ and $dx = 2t dt$.
The integral becomes $I = \int \dfrac{2t}{t^2 + 1 + t} dt = \int \dfrac{2t}{t^2 + t + 1} dt$.
Rewrite the numerator as $2t + 1 - 1$:
$I = \int \dfrac{2t + 1}{t^2 + t + 1} dt - \int \dfrac{1}{t^2 + t + 1} dt$.
The first part is $\ln(t^2 + t + 1)$.
For the second part, complete the square in the denominator: $t^2 + t + 1 = (t + \dfrac{1}{2})^2 + \dfrac{3}{4}$.
Using $\int \dfrac{1}{u^2 + a^2} du = \dfrac{1}{a} \tan^{-1}(\dfrac{u}{a})$, we get $\int \dfrac{1}{(t + 1/2)^2 + (\sqrt{3}/2)^2} dt = \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2t + 1}{\sqrt{3}})$.
Thus, $I = \ln(t^2 + t + 1) - \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2t + 1}{\sqrt{3}}) + C$.
Substituting $t = \sqrt{x-1}$ back, $t^2 + t + 1 = x - 1 + \sqrt{x-1} + 1 = x + \sqrt{x-1}$.
Therefore, $I = \ln(x + \sqrt{x-1}) - \dfrac{2}{\sqrt{3}} \tan^{-1}(\dfrac{2\sqrt{x-1} + 1}{\sqrt{3}}) + C$.
Answer: $\log (x+\sqrt{x-1})-\dfrac{2}{\sqrt{3}} \tan ^{-1}\left(\dfrac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$
Hence, the correct answer is option: 'D'.