Question

Evaluate the following integrals : $$\int \frac{d x}{x^{4}+18 x^{2}+81}$$

Short answer

Answer: The integral of the given function is $$\frac{1}{2} \ln\left|x^2+9\right| + C$$, where $$C$$ is the constant of integration.

Step-by-step solution

Factor the denominator

First, we try to manipulate the denominator to make it look like the sum of squares. We can rewrite the denominator as follows: $$x^4 + 18x^2 + 81 = \left(x^4 + 2 \cdot 3 \cdot 3x^2 + 3^4\right) = \left(x^2 + 9\right)^2$$ Now the denominator is a perfect square, which is what we wanted.

Substitute variables for integration

Let \(u = x^2 + 9\). Then, we will take the derivative, which will give us the following: $$\frac{du}{2} = x\, dx \Rightarrow du = 2x\, dx$$ If we substitute \(x^2 = u - 9\) into the integral, we have: $$\int \frac{1}{(u - 9) + 9}\,dx = \int \frac{1}{u}\,dx$$ But we still need to substitute \(du = 2x\,dx\) to get rid of \(x\). The equation becomes: $$\int \frac{1}{u} \cdot \frac{du}{2x}$$ Now, we cancel out \(x\) in the numerator and the denominator, and we are left with: $$\frac{1}{2} \int \frac{1}{u}\, du$$

Integrate and substitute back

Now, we simply need to evaluate the integral: $$\frac{1}{2} \int \frac{1}{u}\, du = \frac{1}{2} \ln\left| u \right| + C$$ Now, substitute \(u\) back in terms of \(x\): $$\frac{1}{2} \ln\left|x^2+9\right| + C$$ Finally, the integral \(\int \frac{1}{x^4+18x^2+81}\,dx\) evaluates to: $$\frac{1}{2} \ln\left|x^2+9\right| + C$$