$$ \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx = $$
ex√x + C
To solve the integral \(\int e^x \left(\frac{2x + 1}{2\sqrt{x}}\right) \, dx\), we'll use integration by parts, which is a method for integrating products of functions. The formula for integration by parts is given by:
\(\int u \, dv = uv - \int v \, du\)
Let's choose:
- \(u = \frac{2x + 1}{2\sqrt{x}}\)
- \(dv = e^x \, dx\)
\(u = \frac{2x + 1}{2x^{1/2}} = x^{1/2} + \frac{1}{2}x^{-1/2}\)
Differentiate using the power rule:
\(du = \left(\frac{1}{2}x^{-1/2} - \frac{1}{4}x^{-3/2}\right) \, dx\)
\(v = e^x\)
Use the integration by parts formula:
\(\int u \, dv = uv - \int v \, du\)
\(=\left(\frac{2x + 1}{2\sqrt{x}}\right)e^x - \int e^x \left(\frac{1}{2x^{1/2}} - \frac{1}{4x^{3/2}}\right) \, dx\)
Now compute the integral:
\(\int e^x \left(\frac{1}{2x^{1/2}}\right) \, dx = e^x √x + C\)
Important Questions from Integrals
\( \int_{0}^{\frac{\pi}{2}} \frac{1 - \cot x}{\cosec x + \cos x} dx = \)
The value of the integral \( \int_{\log_e 2}^{\log_e 3} \frac{e^{2x}- 1}{e^{2x} + 1} dx \) is :
\(\displaystyle \int \frac{\pi}{x^{n+1} - x} dx\)
\(\int_{2}^{3} |2x - 1| \,dx =\)
\(\int \frac{dx}{x^a} =\)
