Computing the integral $\int \frac{n^x}{n^{2x} + 8n^x + 12}dx$

Question

Compute $$\int \frac{n^x}{n^{2x} + 8n^x + 12} \, dx$$ where $n$ is a positive real number.

I have tried factoring the denominator, but I'm not sure if that takes me anywhere. Does someone know how to do this?

Answer 1

Hint:Take $u=n^x \to du=n^x.\ln n$ and $$\int \frac{n^x}{n^{2x} + 8n^x + 12} \, dx=\int \frac{\dfrac{1}{\ln n}}{u^2+8u+12} \, du=\\\dfrac{1}{\ln n}\int \frac{1}{u^2+8u+12} \, du$$

Answer 2

HINT: Try substitution $u=n^x$

Answer 3

We have $$ I =\int \frac {n^x}{n^{2x}+8n^x+12} dx =\int \frac {n^x}{(n^x+2)(n^x+6)} dx $$ Now substituting $u=\frac {1}{n^x+6} $, we get, $du =-\frac {\ln n n^x}{(n^x+6)^2} dx $ and using $n^x =\frac {1}{u}-6$, we thus have, $$I =\frac{1}{\ln n} \int \frac {1}{4u-1} du $$ Hope you can take it from here.

Answer 4

Substitute $u = n^x$.
We have $du = \log{n }\cdot n^x dx$.
The integral in terms of $u$ is -
$$\frac{1}{\log{n}} \int \frac{1}{u^2 + 8u + 12} du$$ $$= \frac{1}{\log{n}} \int \frac{1}{(u+4)^2 -4} du$$ $$= \frac{1}{4 \cdot \log{n}} \log \bigg \lvert\frac {u+2}{u+6} \bigg \rvert$$