Solve the integral: ln(sqrt(x^2 – 4)) dx

Solve the integral: ln(sqrt(x^2 – 4)) dx

The Correct Answer and Explanation is:

To solve the integral ∫ln⁡(x2−4) dx\int \ln(\sqrt{x^2 – 4}) \, dx∫ln(x2−4​)dx, we can follow a systematic approach.

Step 1: Simplifying the integrand

The first step is to simplify the expression inside the logarithm. We know that:ln⁡(x2−4)=12ln⁡(x2−4)\ln(\sqrt{x^2 – 4}) = \frac{1}{2} \ln(x^2 – 4)ln(x2−4​)=21​ln(x2−4)

So, the integral becomes:∫ln⁡(x2−4) dx=12∫ln⁡(x2−4) dx\int \ln(\sqrt{x^2 – 4}) \, dx = \frac{1}{2} \int \ln(x^2 – 4) \, dx∫ln(x2−4​)dx=21​∫ln(x2−4)dx

Step 2: Integration by parts

Next, we apply the technique of integration by parts. Recall that:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu

Let:

  • u=ln⁡(x2−4)u = \ln(x^2 – 4)u=ln(x2−4), so that du=2xx2−4 dxdu = \frac{2x}{x^2 – 4} \, dxdu=x2−42x​dx
  • dv=dxdv = dxdv=dx, so that v=xv = xv=x

Now, applying the integration by parts formula:∫ln⁡(x2−4) dx=xln⁡(x2−4)−∫2x2x2−4 dx\int \ln(x^2 – 4) \, dx = x \ln(x^2 – 4) – \int \frac{2x^2}{x^2 – 4} \, dx∫ln(x2−4)dx=xln(x2−4)−∫x2−42×2​dx

Step 3: Simplifying the remaining integral

Now, simplify the remaining integral:∫2x2x2−4 dx\int \frac{2x^2}{x^2 – 4} \, dx∫x2−42×2​dx

We can use the fact that:2x2x2−4=2+8×2−4\frac{2x^2}{x^2 – 4} = 2 + \frac{8}{x^2 – 4}x2−42×2​=2+x2−48​

So the integral becomes:∫2x2x2−4 dx=2x+8∫1×2−4 dx\int \frac{2x^2}{x^2 – 4} \, dx = 2x + 8 \int \frac{1}{x^2 – 4} \, dx∫x2−42×2​dx=2x+8∫x2−41​dx

The integral of 1×2−4\frac{1}{x^2 – 4}x2−41​ is a standard one, which is:∫1×2−4 dx=14ln⁡∣x−2x+2∣\int \frac{1}{x^2 – 4} \, dx = \frac{1}{4} \ln \left| \frac{x – 2}{x + 2} \right|∫x2−41​dx=41​ln​x+2x−2​​

So the original integral becomes:∫ln⁡(x2−4) dx=12(xln⁡(x2−4)−2x−2⋅2ln⁡∣x−2x+2∣+C)\int \ln(\sqrt{x^2 – 4}) \, dx = \frac{1}{2} \left( x \ln(x^2 – 4) – 2x – 2 \cdot 2 \ln \left| \frac{x – 2}{x + 2} \right| + C \right)∫ln(x2−4​)dx=21​(xln(x2−4)−2x−2⋅2ln​x+2x−2​​+C)

Step 4: Final answer

Simplifying the expression:∫ln⁡(x2−4) dx=x2ln⁡(x2−4)−x−2ln⁡∣x−2x+2∣+C\int \ln(\sqrt{x^2 – 4}) \, dx = \frac{x}{2} \ln(x^2 – 4) – x – 2 \ln \left| \frac{x – 2}{x + 2} \right| + C∫ln(x2−4​)dx=2x​ln(x2−4)−x−2ln​x+2x−2​​+C

This is the final solution to the integral.

Conclusion

By using integration by parts and simplifying the resulting terms, we arrive at the final expression for the integral of ln⁡(x2−4)\ln(\sqrt{x^2 – 4})ln(x2−4​). This technique relies on reducing the integrand to simpler components and utilizing standard integral formulas.

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