Question
In Exercises $41-44,$ verify by differentiation.$$\int(a x+b)^{n} d x=\frac{1}{a(n+1)}(a x+b)^{n+1}+C \quad(\text { for } n \neq-1)$$
Step 1
The power rule states that the derivative of $x^n$ is $n*x^{n-1}$. Applying this rule, we get: $$ \frac{d}{dx}\left[\frac{1}{a(n+1)}(a x+b)^{n+1}+C\right] = \frac{1}{a(n+1)}(n+1)(a x+b)^{n}(a) + 0 $$ Show more…
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