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?
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – CrSb0001 Jun 06 '23 at 18:26
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I am asking how to integrate this expression without any limits. – DEB SANKAR ROY Jun 06 '23 at 18:27
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Keep factoring, I guess. Although $\ln(1+x^2)$ is somewhat tough. – Sassatelli Giulio Jun 06 '23 at 18:32
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Use integration by parts – MathStackexchangeIsNotSoBad Jun 06 '23 at 18:33
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1$$1+x+x^2+x^3=(1+x)(1+x^2).$$ Integrate by parts: $$\int\ln(1+x)dx=[x\ln(1+x)]-\int\frac x{1+x}dx=[(x+1)\ln(1+x)-x]$$ and $$\int\ln(1+x^2)dx=[x\ln(1+x^2)]-2\int\frac{x^2}{1+x^2}dx=[x\ln(1+x^2)-2x+2\arctan x].$$ – Anne Bauval Jun 06 '23 at 21:47
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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.
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