Question

If \[ \int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x})\,dx = e^{\sqrt{x}}\left[Ax+B\sqrt{x}+C\right]+K, \] then \(A+B+C=\)

• A. \(-2\)
• B. \(2\)
• C. \(4\)
• D. \(-4\)

Hint:

Whenever an integral contains \(\sqrt{x}\) and \(\dfrac{1}{\sqrt{x}}\), substitution \(t=\sqrt{x}\) usually simplifies the integral quickly.

Answer 1

The Correct Option is B

Step 1: Use substitution.
Let \[ \sqrt{x}=t \] Then, \[ x=t^2 \] and \[ dx=2t\,dt \] The given integral becomes \[ \int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x})\,dx \] Substituting \(\sqrt{x}=t\), \(x=t^2\), and \(dx=2t\,dt\), we get \[ \int \frac{e^t}{t}(t^2+t)(2t\,dt) \] \[ = 2\int e^t(t^2+t)\,dt \]

Step 2: Integrate the expression.
Now, \[ 2\int e^t(t^2+t)\,dt = 2\left[\int e^t t^2\,dt+\int e^t t\,dt\right] \] Using the standard results, \[ \int e^t t^2\,dt=e^t(t^2-2t+2) \] and \[ \int e^t t\,dt=e^t(t-1) \] Therefore, \[ 2\int e^t(t^2+t)\,dt = 2e^t\left[(t^2-2t+2)+(t-1)\right] \] \[ = 2e^t(t^2-t+1) \] \[ = e^t(2t^2-2t+2) \]

Step 3: Substitute back \(t=\sqrt{x}\).
Since \[ t=\sqrt{x} \] and \[ t^2=x, \] we get \[ e^t(2t^2-2t+2) = e^{\sqrt{x}}(2x-2\sqrt{x}+2) \] Thus, \[ \int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x})\,dx = e^{\sqrt{x}}(2x-2\sqrt{x}+2)+K \]

Step 4: Compare with the given form.
Given, \[ e^{\sqrt{x}}\left[Ax+B\sqrt{x}+C\right]+K \] Comparing, \[ A=2,\qquad B=-2,\qquad C=2 \] Therefore, \[ A+B+C=2-2+2 \] \[ =2 \]

Step 5: Final conclusion.
Hence, \[ \boxed{2} \]