<s>[INST] How to evaluate this integral $\int \frac{dx}{16+5^{2-3x}}$? 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
 [/INST] Yes, you can use single substitution
Let $16+5^{2-3x}=t\implies 5^{2-3x}\ln 5\cdot (-3)=dt$, $dx=\frac{dt}{3(16-t)\ln 5}$
$$\int \frac{dx}{16+5^{2-3x}}=\int \frac{1}{t}\frac{dt}{3(16-t)\ln 5}$$
$$=\frac{1}{3\ln5}\int \frac{dt}{t(16-t)}$$
$$=\frac{1}{3\ln5}\frac{1}{16}\int\left( \frac{1}{t}-\frac{1}{t-16}\right)\ dt$$
$$=\frac{1}{48\ln5}\int \left(\frac{1}{t}-\frac{1}{t-16}\right)\ dt$$
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