Question

Find the value of the integral \(\displaystyle\int\dfrac{x^{4}}{(x-1)(x^{2}+1)}\,dx\).

Hint:

Divide first since the numerator's degree exceeds the denominator's, then use partial fractions on the proper-fraction remainder.

Answer 1

Step 1: Performing polynomial long division first:
Since the numerator's degree (4) exceeds the denominator's degree (3), divide: \(\dfrac{x^{4}}{(x-1)(x^{2}+1)}=\dfrac{x^4}{x^3-x^2+x-1}=x+1+\dfrac{1}{(x-1)(x^{2}+1)}\) (verified by multiplying back out).

Step 2: Setting up partial fractions for the remainder:
\(\dfrac{1}{(x-1)(x^{2}+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+1}\).

Step 3: Solving for A, B, C:
\(1=A(x^{2}+1)+(Bx+C)(x-1)\). At \(x=1\): \(1=2A\Rightarrow A=\tfrac12\). Matching \(x^2\) coefficients: \(0=A+B\Rightarrow B=-\tfrac12\). Matching constants: \(1=A-C\Rightarrow C=A-1=-\tfrac12\).

Step 4: Writing the full partial fraction decomposition:
\(\dfrac{1}{(x-1)(x^2+1)}=\dfrac{1}{2(x-1)}-\dfrac{x+1}{2(x^{2}+1)}\).

Step 5: Integrating term by term:
\(\displaystyle\int(x+1)dx=\dfrac{x^2}{2}+x\). \(\displaystyle\int\dfrac{dx}{2(x-1)}=\dfrac12\ln|x-1|\). \(\displaystyle\int\dfrac{x+1}{2(x^2+1)}dx=\dfrac14\ln(x^2+1)+\dfrac12\tan^{-1}x\) (splitting \(x/(x^2+1)\) and \(1/(x^2+1)\) separately).

Final Answer:
\[ \boxed{\dfrac{x^{2}}{2}+x+\dfrac12\ln|x-1|-\dfrac14\ln(x^{2}+1)-\dfrac12\tan^{-1}x+C} \]