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Find the general antiderivative of g(x)

Consider the function g(x)=(2x+1)0+6sin⁡(x)−3.g(x) = (2x+1)^0 + 6\sin(x) - 3. First simplify g(x)g(x) as much as possible, then find the general antiderivative G(x)G(x) such that G′(x)=g(x)G'(x) = g(x).

Hint

Remember that anything nonzero raised to the power 0 equals 1. Simplify the constant terms first, then antidifferentiate each piece separately using the rules for constants and sin⁡(x)\sin(x). Don't forget the arbitrary constant.

Step-by-step solution

Since (2x+1)0=1(2x+1)^0 = 1 for all xx where 2x+1≠02x+1 \neq 0, the function simplifies to g(x)=1+6sin⁡(x)−3=6sin⁡(x)−2g(x) = 1 + 6\sin(x) - 3 = 6\sin(x) - 2.

Now antidifferentiate term by term: the antiderivative of 6sin⁡(x)6\sin(x) is −6cos⁡(x)-6\cos(x), and the antiderivative of −2-2 is −2x-2x. Combining these and adding the constant of integration gives G(x)=−6cos⁡(x)−2x+CG(x) = -6\cos(x) - 2x + C.

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