Question

How to solve: Int(ArcTan(x^3)/x^4)dx
Answer:-ArcTan(x^3)/(3*x^3) + Log(x) - Log(1 + x^6)/6

Answer 1

\[
I(n)=\int{\arctan x^{n-1}\over x^n}dx\\
=\int x^{-n}\arctan x^{n-1} dx\\
=x^{-n}\int\arctan x^{n-1} dx-\int\left[{d\over dx}(x^{-n})\int\arctan x^{n-1} dx\right]dx
\]
now,
\[
J(n)=\int\arctan x^{n-1}\cdot1dx\\
J(n)=\arctan x^{n-1}\int1dx-\int\left[{d\over dx}(\arctan x^{n-1})\int1dx\right]dx\\
J(n)=x\arctan x^{n-1}-\int\frac{(n-1)x^n}{x^2(x^n+1)}xdx\\
J(n)=x\arctan x^{n-1}-(n-1)\int\frac{x^{n-1}}{x^n+1}dx\\
J(n)=x\arctan x^{n-1}-(n-1)\frac{\ln(x^n+1)}{n}+C
\]

going back up,
\[
I(n)=x^{-n}J(n)+n\int x^{-n-1}J(n)dx\\
I(n)={J(n)\over x^n}+n\int x^{-(n+1)}[x\arctan x^{n-1}-{n-1\over n}\ln(x^n+1)]dx\\
I(n)={J(n)\over x^n}+n\int{\arctan x^{n-1}\over x^n}dx-(n-1)\int{\ln(x^n+1)\over x^{n+1}}dx\\
2I(n)={J(n)\over x^n}+{n-1\over n}\left[{\ln(x^{-n}+1)}+{\ln(x^n+1)\over x^n}\right]
\]

Answer 2

further simplifying gives
\[
I(n)={x\arctan x^{n-1}\over 2x^n}+{n-1\over 2n}\ln(x^{-n}+1)
\]

Answer 3

problem no.1 , why is arctan -ve??

Answer 4

@rosho could you check for the signs in there?

Answer 5

\[
I(n)={\arctan x^{n-1}\over 2x^{n-1}}+{n-1\over2n}\left[\ln(1+x^n)-\ln(x^n)\right]
\]
this is exactly same as the provided answer but with exactly opposite sign!!!!

Answer 6

oh .. found the correction... the very last step where I substituted J(n) into I(n)