Correct Answer is : 2x−2−22tan⁡−1(x−22)+c2 {x-2}-2 {2} ^{-1}≤ft({{x-2}}{{2}})+c2x−2−22tan−1(2x−2)+c

2 {x-2}-2 {2} ^{-1}≤ft({{x-2}}{{2}})+c

2 {x-2}+2 {2} ^{-1}≤ft({{x-2}}{{2}})+c

2 {x-2}-2 {2} ^{-1}≤ft({{x-2}}{2})+c

2 {x-2}+2 {2} ^{-1}≤ft({{x-2}}{2})+c

Correct Answer is : 2x−2−22tan⁡−1(x−22)+c2 {x-2}-2 {2} ^{-1}≤ft({{x-2}}{{2}})+c2x−2−22tan−1(2x−2)+c

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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{\sqrt{x-2}}{x} d x=

Solution:

Concept:

Use substitution to eliminate the square root, then apply the standard inverse tangent integral formula.

Explanation:

Let

\sqrt{x-2}=t

Then

x-2=t^{2}

, so

x=t^{2}+2

Differentiating,

dx=2t\,dt

Substitute into the integral:

\int \frac{\sqrt{x-2}}{x}\,dx=\int \frac{t}{t^{2}+2}\cdot 2t\,dt=2\int \frac{t^{2}}{t^{2}+2}\,dt.

Write

t^{2}=(t^{2}+2)-2

Therefore,

2\int \frac{t^{2}}{t^{2}+2}\,dt=2\int \left(1-\frac{2}{t^{2}+2}\right)dt=2t-4\int \frac{dt}{t^{2}+2}.

Use the formula

\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)

with

a=\sqrt{2}

So,

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

Hence,

2t-4\cdot \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+c=2t-2\sqrt{2}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+c.

Substitute back

t=\sqrt{x-2}

\int \frac{\sqrt{x-2}}{x}\,dx=2\sqrt{x-2}-2\sqrt{2}\tan^{-1}\left(\frac{\sqrt{x-2}}{\sqrt{2}}\right)+c.

Answer:

Option A:

2\sqrt{x-2}-2\sqrt{2}\tan^{-1}\left(\frac{\sqrt{x-2}}{\sqrt{2}}\right)+c

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