Question

\[\int \frac{x^4}{(x-1)(x^2+1)}dx\]

Answer 1

Divide it out, then use partial fractions

\[\rightarrow \int\limits_{}^{} x +1+\frac{1}{2(x-1)} -\frac{x+1}{2(x^{2}+1)} dx\]

Answer 2

we have
\[\int\limits_{}^{} (x^4)/(x-1)(x^2+1) dx\]
here degree of numerator is greater than denominator
so we'll transform into the form p(x)+q(x)/r(x)
\[\int\limits_{}^{} (x^2(x^2-1)+x^2)/((x-1)(x^2+1)) dx\]
\[\int\limits_{}^{} (x^2/(x-1)- (x^2+1-1)/((x-1)(x^2+1)\]
\[\int\limits_{}^{} x^2/(x-1)-1/(x-1)+1/((x-1)(x^2+1)) dx\]
combining the first two terms
\[\int\limits_{}^{} (x+1)+ 1/((x-1)(x^2+1)) dx\]
split the second term using partial fractions
\[1/(x-1)(x^2+1)= (-x-1)/2(x^2+1)+ 1/2(x-1)\]
so the integral is
\[\int\limits_{}^{} (x+1)+(-x-1)/2(x^2+1) -1/2(x-1) dx\]
\[\int\limits_{}^{} (x+1)-x/2(x^2+1)-1/2(x^2+1)-1/2(x-1) dx\]
now integrating
\[(x+1)^2/2-1/4 \ln(x^2+1)-1/2\tan^{-1} x-1/2 \ln(x-1) + c\]
c= constant of integration

Answer 3

\[=\frac{1}{2}x^{2}+x+\frac{1}{2}\ln(x−1)+\frac{1}{4}\ln(x2+1)+\frac{1}{2}\tan^{−1}x+C\]

Answer 4

This was too easy Ishaan, you should post another one! ;)