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I have this integral to evaluate: $\int 7^{2x+3} dx$

u substitution should work and you are left with $\frac{1}{2}\int7^udx$

And the final answer should be: $\frac{7^{2x+3}}{2\ln7}$

I wasn't too sure about this... so did I do this correctly?

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    I rolled Jack's edit back, as making the correction *dx* -> *du* orphans comments/answers. Also, not sure why people are being stingy with upvotes. This is a well-posed question.2011-12-21

4 Answers 4

-1

You have, although after substitution it should be du not dx but I assume it is a typo and you should have a constant as well.

4

You do not show detail, but you presumably found $\int 7^u\,du$ by looking up a formula for $\int a^u\,du$. That is perfectly correct. But I would use a slightly different approach, which is a bit more complicated but does not rely on remembering $\int a^u\,du$.

Note that $7^{2x+3}=(e^{\ln 7})^{2x+3}=e^{(\ln7)(2x+3)}. \qquad\qquad (\ast)$ Let $v=(\ln 7)(2x+3)$. Then $dv=(\ln 7)(2) \,dx$. Substituting, we find that $\int e^{(\ln7)(2x+3)}\,dx=\int \frac{1}{2\ln 7}e^v\,dv=\frac{1}{2\ln 7}e^v+C.$ Finally, by $(\ast)$, $e^v=7^{2x+3}$ so our integral is $\frac{1}{2\ln 7}7^{2x+3}+C.$

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    :) ${}{}{}{}{}$2011-12-21
2

You can always differentiate to be sure:

$ \frac{\mathrm{d}}{\mathrm{d}x} \frac{7^{2x+3}}{2\ln 7} = \frac1{2 \ln 7} \frac{\mathrm{d}}{\mathrm{d}x} 7^{2x+3} = \frac1{2 \ln 7} (\ln 7) 7^{2x+3} \frac{\mathrm{d}}{\mathrm{d}x} (2x+3) = 7^{2x+3} $

Your result is sort of right; you're missing the constant and the differential after the substitution should be $\mathrm{d}u$, so the result is the following:

$ \int 7^{2x+3}\ \mathrm{d}x = \frac1{2}\int 7^u \mathrm{d}u =\frac{7^{2x+3}}{2\ln 7} + C $

-1

Yes it is correct.

except for the + Constant

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    And except for the *dx* in place of *du*, and the use of "equation"...2011-12-21