Consider the indefinite integral [tex]$\int \frac{-3 e^{-3 x}}{\left(e^{-3 x}+3\right)^2} d x$[/tex]:
This can be transformed into a basic integral by letting
[tex]$u = \square$[/tex] and
[tex]$d u= \square$[/tex] [tex]$d x$[/tex]
Performing the substitution yields the integral
[tex]$\int \square du$[/tex]
Answer (2)
To evaluate the integral ∫ ( e − 3 x + 3 ) 2 − 3 e − 3 x d x , we use the substitution u = e − 3 x + 3 , which leads to d u = − 3 e − 3 x d x . This transforms the integral into ∫ u 2 1 d u , which integrates to − u 1 + C or − e − 3 x + 3 1 + C .
;
We identify a suitable substitution for the given integral.
Let u = e − 3 x + 3 , then calculate d u = − 3 e − 3 x d x .
Substitute u and d u into the integral, which transforms the integral to ∫ u 2 1 d u .
The final answer is u = e − 3 x + 3 , d u = − 3 e − 3 x d x , and the transformed integral is ∫ u 2 1 d u .
Explanation
Problem Analysis We are given the indefinite integral ∫ ( e − 3 x + 3 ) 2 − 3 e − 3 x d x and we want to find a suitable substitution to simplify it.
Finding du Let's try the substitution u = e − 3 x + 3 . Then, we need to find the differential d u . Taking the derivative of u with respect to x , we have d x d u = − 3 e − 3 x Thus, d u = − 3 e − 3 x d x .
Performing the Substitution Now we can substitute u and d u into the original integral. We have ∫ ( e − 3 x + 3 ) 2 − 3 e − 3 x d x = ∫ u 2 1 d u So the integral becomes ∫ u 2 1 d u .
Final Answer Therefore, the substitution is u = e − 3 x + 3 , d u = − 3 e − 3 x d x , and the transformed integral is ∫ u 2 1 d u .
Examples
In physics, this type of integral can appear when calculating the total charge on a capacitor as it discharges over time. The exponential term represents the decaying current, and the integral helps determine the accumulated charge. Understanding substitutions like this is crucial for solving many problems in electrical engineering and other fields.
