CUET Mathematics 2024 - e^x(2 x+12 x) dx | PYQs + Solutions

CUET Mathematics 2024

Calculus

Integrals

Medium

$\int \mathrm{e}^{\mathrm{x}}\left(\frac{2 \mathrm{x}+1}{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}$

\frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{e}^{\mathrm{x}}+\mathrm{C}

-\mathrm{e}^{\mathrm{x}} \sqrt{\mathrm{x}}+C

-\frac{1}{2 \sqrt{x}} e^{x}+C

e^{x} \sqrt{x}+C

✅ Correct Option: 4

First, let's simplify the expression:

\int e^x \left(\dfrac{2x+1}{2\sqrt{x}}\right) dx = \int e^x \cdot \dfrac{2x+1}{2} \cdot x^{-1/2} dx

= \int e^x \cdot \left(\dfrac{2x}{2} \cdot x^{-1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx

= \int e^x \cdot \left(x^{1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx

Splitting the integral:

\int e^x \cdot \left(x^{1/2} + \dfrac{1}{2} \cdot x^{-1/2}\right) dx = \int e^x \cdot x^{1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx

Using integration by parts for the first part with:

u = x^{1/2}

du = \dfrac{1}{2} \cdot x^{-1/2} dx

dv = e^x dx

v = e^x

\int e^x \cdot x^{1/2} dx = x^{1/2} \cdot e^x - \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx

Notice that the second term is the negative of our second integral:

\int e^x \cdot x^{1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx = x^{1/2} \cdot e^x - \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx + \int e^x \cdot \dfrac{1}{2} \cdot x^{-1/2} dx

The last two integrals cancel out, giving us:

\int e^x \left(\dfrac{2x+1}{2\sqrt{x}}\right) dx = \sqrt{x} \cdot e^x + C