How to evaluate this integral $\int \frac{dx}{16+5^{2-3x}}$?

Question

Evaluate the following integral $$\int \frac{dx}{16+5^{2-3x}}=?$$

My attempt:

Assume that $2-3x=t$, $-3dx=dt$, $dx=-\frac{dt}{3}$ $$\int \frac{-dt/3}{16+5^{t}}=-\frac13\int \frac{dt}{16+5^{t}}$$ let $ 5^t=u$, $\implies 5^t\ln 5\ dt=du$, $dt=\frac{du}{u\ln 5}$ $$-\frac13\int \frac{\frac{du}{u\ln 5}}{16+u}$$ $$=-\frac1{3\ln 5}\int \frac{du}{u(16+u)}$$ $$=-\frac1{3\ln 5}\int \frac{1}{16}(\frac{1}{u}-\frac{1}{16+u})du$$ $$=-\frac1{48\ln 5}(\ln u-\ln(u+16))$$ I back substituted, $u=5^t$ $$=-\frac1{48\ln 5}(\ln 5^t-\ln(5^t+16))$$

I again back substituted, $t=2-3x$ $$=-\frac1{48\ln 5}(\ln 5^{2-3x}-\ln(5^{2-3x}+16))$$ after simplification i got the answer $$=\frac1{48\ln 5}\ln\left(\frac{16+5^{2-3x}}{5^{2-3x}}\right)+c$$ In this integral I used two succesive substitutions, which is a bit lengthier. Can the same integral be calculated by some easier method? Can I evaluate by a single substitution? some body please help me. Thanks

Answer 1

Yes, you can use single substitution

Let $16+5^{2-3x}=t\implies 5^{2-3x}\ln 5\cdot (-3)=dt$, $dx=\frac{dt}{3(16-t)\ln 5}$ $$\int \frac{dx}{16+5^{2-3x}}=\int \frac{1}{t}\frac{dt}{3(16-t)\ln 5}$$ $$=\frac{1}{3\ln5}\int \frac{dt}{t(16-t)}$$ $$=\frac{1}{3\ln5}\frac{1}{16}\int\left( \frac{1}{t}-\frac{1}{t-16}\right)\ dt$$ $$=\frac{1}{48\ln5}\int \left(\frac{1}{t}-\frac{1}{t-16}\right)\ dt$$

Answer 2

It saves a substitution if you do this...

$\int \frac {dx}{16 + 5^{2-3x}}\ dx\\ \int \frac {dx}{16 + 25 (5^{-3x})}\ dx\\ \int \frac {dx}{16 + 25 e^{-3x\ln 5}}\ dx$

Now our substitution $u = e^{-3x\ln 5}, du = e^{-3x\ln 5}(-3\ln 5)\ dx, dx = -\frac {1}{3\ln 5 u}\ du$

$-\frac {1}{3\ln 5}\int \frac {1}{u(16 + 25u)} \ du$

And proceed with the partial fractions as above.

Answer 3

The integral can be written as $$I=\int \frac{5^{3x}dx}{16.5^{3x}+25}$$ with $5^{3x}=e^{3x\ln(5)}$,

The derivative of the denominator is $$16.3.\ln(5).e^{3x\ln(5)}=48\ln(5)5^{3x}$$

thus

$$I=\frac{1}{48\ln(5)}\ln(16.5^{3x}+25)+C$$