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Consider the indefinite integral

6 + e^x

The most appropriate substitution to simplify this integral is u = f(x) where f(x) = e^x+6

We then have

dx = g(u) du

where

g(u) = e^-u

Hint: you need to back substitute for x in terms of u for this part.

After substituting into the original integral we obtain

h(u) du where

h(u) = 1/(1-6/u)^2

To evaluate this integral, rewrite the numerator as

6 = u – (u – 6)

simplify; then integrate, thus obtaining

h(u) du = H(u)

where H(u) = 1/6 * ln(1-6/u)

After substituting back for u, we obtain our final answer: âˆ«(e^x+6) dx = -ln(e^x+6) + 6 + C

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Transcript

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