Question

$\int e^{kx}$ [f(kx) + f'(kx)]dx will be equal to $\frac{1}{k}{e}^{kx}$ f(kx) + C only when k = 1.

• A. True
• B. False

Answer 1

The correct option is B

False

Let's proceed the same way we found the formula for the integral of $\int e^x$ [f(x) + f'(x)]dx

$\int e^{kx}$ [f(kx) + f'(kx)]dx = $\int e^{kx}$ f(kx) + $e^{kx}$ f'(kx)]dx

Now apply integration by parts on the first integral. f(kx) will be the first function and $e^{kx}$ will be second function.

$\frac{e^{kx}.f\left(kx\right)}{k}$ - $\int \frac{e^{kx}.f^{\prime }\left(kx\right).k}{k}$ dx + $\int e^{kx}$ f'(kx)dx

$=\frac{e^{kx}.f\left(kx\right)}{k}-\int e^{kx}$ f'(kx)dx + $\int e^{kx}$ f'(kx)dx + C

$=\frac{e^{kx}.f\left(kx\right)}{k}$ + C

Here, we see that k can take any real value except zero. So the given statement that it's true for only for k = 1 is not true.