The value of \(\rm \int \sqrt{x}e^{\sqrt{x}}\ dx\) is equal to:1. \(

1.	The value of \(\rm \int \sqrt{x}e^{\sqrt{x}}\ dx\) is equal to:1. \(\rm 2\sqrt{x}-e^{\sqrt{x}}-4\sqrt{xe^{\sqrt{x}}}+C\)2. \(\rm (2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\)3. \(\rm (2x+4\sqrt{x}+4)e^{\sqrt{x}}+C\)4. \(\rm (1-4\sqrt{x})e^{\sqrt{x}}+C\)
Answer» Correct Answer - Option 2 : \(\rm (2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\) Concept: Integration by Parts: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx. Integration by substitution: If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt. ∫ e^x dx = e^x + C. Calculation: Let \(\rm I=\int \sqrt{x}e^{\sqrt{x}}\ dx\) Substituting \(\rm \sqrt x=t\), we get: \(\rm \frac{1}{2\sqrt x}dx=dt\) ⇒ dx = 2t dt ∴ \(\rm I=\int t\times e^t\times2t\ dt\) Integrating by parts, taking t² as the first function and e^t as the second function, we get: ⇒ \(\rm I=2\left[t^2 \int e^t\ dt-\int \left(\frac{d}{dt}t^2\int e^t\ dt \right)\ dt \right]+C\) ⇒ \(\rm I=2t^2 e^t-4\int t e^t\ dt+C\) Integrating \(\rm \int t e^t\ dt\) by parts, we get: ⇒ \(\rm I=2t^2 e^t-4\left[t \int e^t\ dt-\int \left(\frac{d}{dt}t\int e^t\ dt \right)\ dt \right]+C\) ⇒ \(\rm I=2t^2 e^t-4\left(t e^t-e^t\right)+C\) ⇒ \(\rm I=\left(2t^2-4t+4\right)e^t+C\) Back substituting \(\rm \sqrt x=t\), we get: ⇒ \(\rm I=(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\).

The value of \(\rm \int \sqrt{x}e^{\sqrt{x}}\ dx\) is equal to:1. \(\rm 2\sqrt{x}-e^{\sqrt{x}}-4\sqrt{xe^{\sqrt{x}}}+C\)2. \(\rm (2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\)3. \(\rm (2x+4\sqrt{x}+4)e^{\sqrt{x}}+C\)4. \(\rm (1-4\sqrt{x})e^{\sqrt{x}}+C\)

Answer» Correct Answer - Option 2 : \(\rm (2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\)

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Integration by substitution:

If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.

Calculation:

Let \(\rm I=\int \sqrt{x}e^{\sqrt{x}}\ dx\)

Substituting \(\rm \sqrt x=t\), we get:

\(\rm \frac{1}{2\sqrt x}dx=dt\)

⇒ dx = 2t dt

∴ \(\rm I=\int t\times e^t\times2t\ dt\)

Integrating by parts, taking t² as the first function and e^t as the second function, we get:

⇒ \(\rm I=2\left[t^2 \int e^t\ dt-\int \left(\frac{d}{dt}t^2\int e^t\ dt \right)\ dt \right]+C\)

⇒ \(\rm I=2t^2 e^t-4\int t e^t\ dt+C\)

Integrating \(\rm \int t e^t\ dt\) by parts, we get:

⇒ \(\rm I=2t^2 e^t-4\left[t \int e^t\ dt-\int \left(\frac{d}{dt}t\int e^t\ dt \right)\ dt \right]+C\)

⇒ \(\rm I=2t^2 e^t-4\left(t e^t-e^t\right)+C\)

⇒ \(\rm I=\left(2t^2-4t+4\right)e^t+C\)

Back substituting \(\rm \sqrt x=t\), we get:

⇒ \(\rm I=(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\).