I am trying to solve this integral which looks simple but for some reason I can't reach the final result.
$$\int\ln(1+x^4) dx$$
I have no idea as to what to try.
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3Start with integration by parts, but this integral is anything but simple! – Tim Thayer Apr 15 '17 at 05:37
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$$\int \ln(1+x^4).\,x=\ln(1+x^4)\cdot x-\int4\frac{x^4}{1+x^4}.dx$$ Now You may further integrate. Hint: To Do $\int \frac 1 {1+x^4}\, dx$ use factorization.
The Dead Legend
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Write $2=1+x^2+1-x^2$ and use the trick of https://math.stackexchange.com/questions/426152/evaluating-int-0-infty-fracdx1x4/426194#426194 or https://math.stackexchange.com/questions/2234797/int-e3x-ex-over-e4x-e2x-1-dx/2234829#2234829 – lab bhattacharjee Apr 15 '17 at 06:13
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Using by parts we have $x\ln (1+x^4)-\int\frac {2x^3.2x}{1+x^4} $. Now use $x^2=u $ thus $2xdx=du $ then the integral part turns to $\int \frac{2u^\frac {3}{2}}{1+u^2} $ now using $u=\tan (t)$ and simplifying we have $\int 2tan^{\frac {3}{2}}(t)dt$ now writiong it as $\frac {\tan^2 (t)}{\sqrt {\tan(t)}}$ we separate $tan^2 (t)=\sec^2 (t)-1$ and now we can perform integration by separating two parts. Hope you know how to integrate $\int \sqrt {cot (t)} dt$
Archis Welankar
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If you are unafraid of complex numbers write $1+x^4$ as $(x-u_1)\cdots(x-u_4)$ where $u_k=\exp(\pi i(2k-1)/4)$. Your integral is the sum of the $\int\ln(x-u_k)\,dx$ and these can be done by the usual trick (answer $(x-u_k)\ln(x-u_k)-(x-u_k)$.) Finally, you need to write out these complex logs in real terms.
Angina Seng
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Yes +1 this would be my answer too. But it's maybe too strange for students in their first calculus class. – mathreadler Apr 15 '17 at 12:19