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int(sqrt(x^(2)+1))/(x^(2)sqrt(x^(2)-1))dx

∫x2+1x2x2−1dx \int \frac{\sqrt{x^{2}+1}}{x^{2} \sqrt{x^{2}-1}} d x

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. ∫x2+1x2x2−1dx \int \frac{\sqrt{x^{2}+1}}{x^{2} \sqrt{x^{2}-1}} d x
  1. Identify Substitution: We start by identifying a substitution. Let u=x2−1u = x^2 - 1, then du=2xdxdu = 2x dx. This substitution simplifies the integral.
  2. Rewrite Integral Using Substitution: Rewrite the integral using the substitution: u=x2−1u = x^2 - 1, so x2=u+1x^2 = u + 1. dx=du2x=du2u+1dx = \frac{du}{2x} = \frac{du}{2\sqrt{u+1}}. The integral becomes ∫u+1+1(u+1)u⋅du2u+1\int \frac{\sqrt{u+1+1}}{(u+1)\sqrt{u}} \cdot \frac{du}{2\sqrt{u+1}}.
  3. Simplify Expression: Simplify the expression: ∫u+2(u+1)u⋅du2u+1=∫u+2((u+1)u⋅2u+1)du\int\frac{\sqrt{u+2}}{(u+1)\sqrt{u}} \cdot \frac{du}{2\sqrt{u+1}} = \int\frac{\sqrt{u+2}}{((u+1)\sqrt{u} \cdot 2\sqrt{u+1})} du.
  4. Further Simplify: Further simplify: ∫u+2((u+1)u⋅2u+1)du=∫u+22u(u+1)32du.\int\frac{\sqrt{u+2}}{((u+1)u \cdot 2\sqrt{u+1})} du = \int\frac{\sqrt{u+2}}{2u(u+1)^{\frac{3}{2}}} du.
  5. Check for Simpler Approach: This integral looks complex and might require special functions or numerical methods for exact evaluation. However, let's check if there's a simpler approach or error in previous steps.

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