Question

how to integrate this with respect to x ?
1/(x((x^n)+1))

Answer 1

\[\int\limits \frac{ dx }{(x(x^n +1))}\]

Answer 2

try below :

\[ \begin{array} \\
\int \dfrac{1}{x(x^n+1)}dx &= \int \dfrac{(x^n+1) -x^n}{x(x^n+1)}dx \\ &= \int \dfrac{(x^n+1) }{x(x^n+1)}dx - \int \dfrac{x^n}{x(x^n+1)}dx

\end{array} \]

Answer 3

ok... i'm trying

Answer 4

done i believe...is this the answer:
\[\ln x +\frac{\ln(x^n +1) }{ n }\]

Answer 5

\[\ln x \color{red}{-}\frac{\ln(x^n +1) }{ n } + C\]

Answer 6

thanks
but guys can you think of some other way too??....

Answer 7

do not go for more high performance. do as much as you can

Answer 8

What I did was pick \[\LARGE u=x^n+1\] which gave me \[\LARGE du=nx^{n-1}dx\] Just rearranging this with algebra gives me: \[\LARGE \frac{du}{nx^n}=\frac{dx}{x}\] Now on the du side I plug in x^n from the original substitution: \[\LARGE \frac{du}{n(u-1)}=\frac{dx}{x}\] Now I can plug in everything to the original equation:

\[\LARGE \int\limits \frac{1}{x^n+1}\frac{dx}{x} = \int\limits \frac{1}{u}\frac{du}{n(u-1)}\] Rearrange a little to get: \[\LARGE \frac{1}{n}\int\limits \frac{du}{u(u-1)}\] Then from here I do partial fractions: \[\LARGE \frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{(u-1)}\] To solve for A and B I make them have a common denominator: \[\LARGE 1=A(u-1)+Bu\] since u is a variable, we can plug in any value. Plug in the values u=1 and u=0 to solve for A and B \[\LARGE A=-1, B=1\] Now plug back into the integral: \[\LARGE \frac{1}{n} \int\limits (\frac{-1}{u}+\frac{1}{u-1})du\] Integrate each one to get: \[\LARGE \frac{-1}{n} \ln u +\frac{1}{n} \ln(u-1)+C\] Plug back in u=x^n+1 \[\LARGE \frac{-1}{n} \ln (x^n+1) +\frac{1}{n} \ln(x^n)+C\] Use log rules on the second logarithm to get: \[\LARGE \ln(x)-\frac{1}{n} \ln (x^n+1) +C\] Hey, we're done. Ok, maybe not as nice as ganeshie's answer but probably slightly easier to stumble upon by accident by trying u-substitution.

I hope this gives you something you wanted! @Mritunjay22 It's always good to be able to find alternate routes to the same answer and try to expand your thinking! Continue to strive for more! =D