How to integrate the above integrand?

Question

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How to integrate the above integrand? We can write the above integrand as $\ln\left(\frac{1-x^4}{1-x}\right)$. So, it will become $\ln(1-x^4)-\ln(1-x)$. But how to solve after that?

Answer 1

Suppose that $x = g(y)$ is the inverse function of $y = f(x)$. Then we have $$\int g(y) \;dy = \int g(f(x)) \cdot f'(x) \;dx = \int x \cdot f'(x)\; dx$$

For example take $f(x) = e^x, g(y) = \log y$, then $x \cdot f'(x) = xe^x$ which is easy to integrate: $$\int xe^x \;dx = (x - 1)e^x + C$$ Now replace $x$ with $\log y$ and we get $$\int \log y \;dy = y \log y - y + C$$ which is well-known.

Now we apply this trick to the integral $\int \log(y^2 + 1) \;dy$. Take $g(y) = \log (y^2 + 1), f(x) = \sqrt{e^x - 1}$. Then we have to evaluate $$\int x \cdot f'(x) \;dx = \int \frac{xe^x}{2\sqrt{e^x - 1}} \;dx$$ The RHS is very easy to evaluate, using integration by parts. Take $u = x, v' = e^x/2\sqrt{e^x - 1}$. To find $v$ make a change of variable $z = e^{x/2}$, so $v' = z' \cdot z/\sqrt{z^2 - 1}$, and $v = \sqrt{z^2 - 1} = \sqrt{e^x - 1}$.

Hence $$\int \frac{xe^x}{2\sqrt{e^x - 1}} = x\sqrt{e^x - 1} - \int \sqrt{e^x - 1} \;dx$$

The remaining integral can be evaluated with Euler substitution, by taking $z = e^{x/2}$: $$\int \sqrt{e^x - 1} \;dx = \int \frac{e^{x/2}}{2} \cdot \frac{2\sqrt{e^x - 1}}{e^{x/2}} \; dx = 2\int \frac{\sqrt{z^2 - 1}}{z} \;dz = 2\sqrt{z^2 - 1} - 2\arctan{\sqrt{z^2 - 1}} + C$$

Hence

$$\int \frac{xe^x}{2\sqrt{e^x - 1}} = (x - 2)\sqrt{e^x - 1} + 2\arctan{\sqrt{e^x - 1}} + C$$

Substituting back $x = \log(y^2 + 1)$, we get $$\int \log(y^2 + 1) \;dy = y\log(y^2 + 1) - 2y + 2\arctan(y) + C$$

Since $1 + x + x^2 + x^3 = (x^2 + 1)(x + 1)$, you can combine the above result with other well-known results to evaluate your integral. It is also possible to derive the above result by working directly on $\log(x^2 + 1)$, but it would be harder to see how to choose $u,v$ in integration by parts.