AP EAMCET · Maths · Indefinite Integration
If \(I_{m, n}=\int e^{m x} \cdot x^n d x\), then \(I_{m, n}+\frac{n}{m} I_{m, n-1}=\)
- A \(x^n \cdot e^{m x}+c\)
- B \(\frac{x^n e^{m x}}{n}+c\)
- C \(\frac{x^n \cdot e^{m x}}{m}+c\)
- D \(\frac{-x^n \cdot e^{m x}}{m}+c\)
Answer & Solution
Correct Answer
(C) \(\frac{x^n \cdot e^{m x}}{m}+c\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \because I_{m, n}=\int e^{m x} \cdot x^n d x \\ & =\frac{1}{m} x^n e^{m x}-\int\left(n x^{n-1}\right)\left(\frac{e^{m x}}{m}\right) d x \\ & =\frac{x^n e^{m x}}{m}-\frac{n}{m} \int e^{m x} x^{n-1} d x=\frac{x^n e^{m x}}{m}-\frac{n}{m} I_{m, n-1}+c \\ &…
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