# Evaluating $\int 7^{2x+3} dx$

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: $$\frac{7^{2x+3}}{2\ln7}$$

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Try differentiating this expression. You are solving for an indefinite integral, the definition of which is an expression that when differentiated, yields the original. – EuYu Dec 21 '11 at 1:48
Little mathematical English lesson: if you see no "$=$" anywhere in what you have, then that cannot be called an equation. Also, one does not solve for integrals, since again there is no "$=$" anywhere. Integrals are evaluated, not solved. – J. M. Dec 21 '11 at 2:36
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. – The Chaz 2.0 Dec 21 '11 at 16:17

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:

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

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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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Thank you. It is is nice seeing how to actually get the answer instead of just remembering the formula like you said. – SineCosine Dec 21 '11 at 4:13
I can't see the first mention of (star) on my phone! – The Chaz 2.0 Dec 21 '11 at 4:32
@The Chaz: So you can see stars through your phone! Mine has only pulse dialing, is heavy, and black. (I had left out the first star, thanks!) – André Nicolas Dec 21 '11 at 4:56
:) ${}{}{}{}{}$ – The Chaz 2.0 Dec 21 '11 at 5:47

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"... – The Chaz 2.0 Dec 21 '11 at 1:54

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.

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