Question

If \( \int e^{5x}x^{n} \, dx = F(n, x) + c \), then \( 5F(n, x) + nF(n-1, x) = \)

• A. \( F'(n, x) + k \)
• B. \( -F'(n, x) + k \)
• C. \( \frac{xF'(n, x)}{5} + k \)
• D. \( \frac{x^{2}F'(n, x)}{F(n, x)} + k \)

Hint:

Reduction formulas are just structural rearrangements of integration by parts. Recognizing that the derivative of the integral equals the integrand lets you skip solving the reduction loop entirely.

Answer 1

The Correct Option is A

Step 1: Note what the derivative gives.
Since $F(n,x)=\int e^{5x}x^n\,dx$, differentiating undoes the integral: \[ F'(n,x)=e^{5x}x^n \]
Step 2: Integrate by parts.
For $\int e^{5x}x^n\,dx$, take $u=x^n$ and $dv=e^{5x}dx$, so $du=nx^{n-1}dx$ and $v=\frac{e^{5x}}{5}$.
Step 3: Write the reduction relation.
\[ F(n,x)=\frac{1}{5}e^{5x}x^n-\frac{n}{5}\int e^{5x}x^{n-1}dx=\frac{1}{5}e^{5x}x^n-\frac{n}{5}F(n-1,x) \]
Step 4: Multiply through by 5.
\[ 5F(n,x)=e^{5x}x^n-nF(n-1,x) \]
Step 5: Rearrange.
\[ 5F(n,x)+nF(n-1,x)=e^{5x}x^n \]
Step 6: Replace using the derivative.
Since $e^{5x}x^n=F'(n,x)$, \[ 5F(n,x)+nF(n-1,x)=F'(n,x) \] \[ \boxed{F'(n,x)+k} \] (the $k$ allows for the constant of integration).