Question

If $\int e^{2x}f' \left(x\right)dx =g \left(x\right)$, then $\int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$

• A. $\frac{1}{2} [e^{2x} f(x) - g (x)] + C$
• B. $\frac{1}{2} [e^{2x} f(x) + g (x)] + C$
• C. $\frac{1}{2} [e^{2x} f'(2x) + g (x)] + C$
• D. $\frac{1}{2} [e^{2x} f'(2x) + g (x)] + C$

Answer 1

Answer: (A)

$\frac{1}{2} [e^{2x} f(x) - g (x)] + C$

Answer 2

We have,
$\int e^{2 x} f^{\prime}(x) d x=g(x)$
Let $I=\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right) d x$
$\left.=f(x) \int e^{2 x} d x-\int f^{\prime}(x) \int e^{2 x} d x\right) d x+\int e^{2 x} f^{\prime}(x) d x$
$=\frac{f(x) e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x+\int e^{2 x} f^{\prime}(x) d x$
$=\frac{e^{2 x}}{2} f(x)-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x$
$=\frac{1}{2}\left[e^{2 x} f(x)-\int e^{2 x} f^{\prime}(x) d x\right]$
$=\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C$