Problem 73
Medium Rational and Quadratic

Evaluate the integral: integral (x dx)/(5 + x^4)

Evaluate the integral: $$ \int \frac{x\,dx}{5 + x^4}. $$

Explanation of Method

Since $x^4=(x^2)^2$, substituting $u=x^2$ turns the denominator into $5+u^2$, a standard arctangent form, while $x\,dx$ converts neatly into $\tfrac12 du$.

  1. Substitute $u=x^2$ to expose the arctangent structure.
  2. Convert $x\,dx$ into a constant multiple of $du$.
  3. Apply the standard arctangent integral formula.
  4. Substitute back and add the constant of integration.

Step 1 · Substitution

We solve this integral using the substitution method. Let:

$$ u = x^2. $$

Then, differentiate both sides with respect to $x$:

$$ \frac{du}{dx} = 2x, \quad \text{so} \quad du = 2x\,dx. $$

Therefore, $x\,dx = \dfrac{du}{2}$.

Step 2 · Substitute into the integral

Substitute this into the original integral:

$$ \int \frac{x\,dx}{5+x^4} = \int \frac{\frac{du}{2}}{5+u^2} = \frac{1}{2}\int \frac{du}{5+u^2}. $$

Step 3 · Apply the standard result

We recognize that the integral $\displaystyle \int \frac{du}{a^2+u^2}$ is a standard integral, whose solution is:

$$ \frac{1}{a}\arctan\left(\frac{u}{a}\right). $$

In our case, $a^2=5$, so $a=\sqrt5$. Thus, we have:

$$ \frac{1}{2}\int \frac{du}{5+u^2} = \frac{1}{2}\cdot \frac{1}{\sqrt5}\arctan\left(\frac{u}{\sqrt5}\right). $$

Step 4 · Substitute back

Now, substitute $u=x^2$ back into the solution:

$$ \frac{1}{2\sqrt{5}}\arctan\left(\frac{x^2}{\sqrt{5}}\right) + C. $$


Summary Table

Quantity Value
Substitution$u = x^2$
$x\,dx$ in terms of $du$$\tfrac12 du$
Antiderivative$\tfrac{1}{2\sqrt5}\arctan(u/\sqrt5)$

Additional Notes

Result

The solution to the integral is:

$$ \frac{1}{2\sqrt{5}}\arctan\left(\frac{x^2}{\sqrt{5}}\right) + C. $$