Evaluate the integral: I = integral (x^5+1)/(x^4(x²+1)) dx

Evaluate the integral: $$ I = \int \frac{x^5+1}{x^4(x^2+1)}\,dx. $$

Step 1 · Split the integrand

$$ \frac{x^5+1}{x^4(x^2+1)} = \frac{x}{x^2+1} + \frac{1}{x^4(x^2+1)}. $$

Step 2 · Integrate the first piece

$$ \int \frac{x}{x^2+1}\,dx = \frac12\ln\left(x^2+1\right). $$

Step 3 · Partial fractions for the second piece

Since the integrand is even, write $\dfrac{1}{x^4(x^2+1)}=\dfrac{B}{x^2}+\dfrac{D}{x^4}+\dfrac{F}{x^2+1}$. Multiplying through: $1=Bx^2(x^2+1)+D(x^2+1)+Fx^4$. Matching coefficients gives $D=1$, $B=-1$, $F=1$.

Step 4 · Integrate the second piece

$$ \frac{1}{x^4(x^2+1)} = -\frac{1}{x^2} + \frac{1}{x^4} + \frac{1}{x^2+1}, $$

$$ \int \frac{1}{x^4(x^2+1)}\,dx = \frac1x - \frac{1}{3x^3} + \arctan x. $$

Step 5 · Combine

$$ I = \frac12\ln\left(x^2+1\right) + \frac1x - \frac{1}{3x^3} + \arctan x + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$, where the denominator vanishes.
• Since $1/[x^4(x^2+1)]$ is an even function, no odd-power terms ($1/x$, $1/x^3$, or an $x/(x^2+1)$ piece) can appear in its own partial-fraction decomposition — only $1/x^2$, $1/x^4$, and $1/(x^2+1)$ survive, which simplifies the coefficient-matching considerably.
• The result can be checked by differentiating $\tfrac1x-\tfrac1{3x^3}+\arctan x$: this gives $-\tfrac1{x^2}+\tfrac1{x^4}+\tfrac1{x^2+1}$, matching the partial-fraction decomposition; adding back the derivative of $\tfrac12\ln(x^2+1)$, namely $x/(x^2+1)$, and combining everything over $x^4(x^2+1)$ reduces to exactly $(x^5+1)/[x^4(x^2+1)]$, matching the original integrand.