Question:
If \(x > 0\), then \(\int \frac{1}{\sqrt{x^4 + 2x^3 + 2x^2}} dx =\)
If \(x > 0\), then \(\int \frac{1}{\sqrt{x^4 + 2x^3 + 2x^2}} dx =\)
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When the integrand has the form \(\int \frac{1}{x \sqrt{ax^2 + bx + c}} dx\), the substitution \(x = 1/t\) is a powerful technique to convert it into a standard integral involving a square root in the denominator.
Updated On: Jun 9, 2026
- \( -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x+2}{x}\right) + c \)
- \( \frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x+2}{x}\right) + c \)
- \( -\frac{1}{\sqrt{2}} \cosh^{-1}\left(\frac{x+2}{x}\right) + c \)
- \( \frac{1}{\sqrt{2}} \cosh^{-1}\left(\frac{x+2}{x}\right) + c \)
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The Correct Option is A
Solution and Explanation
Concept:
To solve this integral, we simplify the expression inside the square root and use appropriate algebraic manipulations or substitutions to transform it into a standard integral form.
Step 1: Simplify the expression inside the square root.
Factor out \(x^2\) from the square root: \[ \sqrt{x^4 + 2x^3 + 2x^2} = \sqrt{x^2(x^2 + 2x + 2)} = x\sqrt{x^2 + 2x + 2} \] So the integral becomes: \[ I = \int \frac{1}{x\sqrt{x^2 + 2x + 2}} \, dx \]
Step 2: Apply substitution to simplify the integrand.
Let \(x = \frac{1}{t}\), then \(dx = -\frac{1}{t^2} dt\). Substituting these into the integral: \[ I = \int \frac{1}{(1/t)\sqrt{(1/t)^2 + 2(1/t) + 2}} \left(-\frac{1}{t^2}\right) \, dt \] \[ I = -\int \frac{1}{t\sqrt{\frac{1 + 2t + 2t^2}{t^2}}} \, dt = -\int \frac{1}{\sqrt{2t^2 + 2t + 1}} \, dt \]
Step 3: Complete the square and integrate.
Factor out \(\sqrt{2}\) from the square root: \[ I = -\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{t^2 + t + 1/2}} \, dt \] Completing the square for \(t^2 + t + 1/2 = (t + 1/2)^2 + 1/4\): \[ I = -\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{(t + 1/2)^2 + (1/2)^2}} \, dt \] Using the standard integral \(\int \frac{1}{\sqrt{u^2 + a^2}} du = \sinh^{-1}(u/a)\): \[ I = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{t + 1/2}{1/2}\right) + c = -\frac{1}{\sqrt{2}} \sinh^{-1}(2t + 1) + c \] Since \(t = 1/x\): \[ I = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{2}{x} + 1\right) + c = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x + 2}{x}\right) + c \]
Conclusion: The value of the integral is \( -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x+2}{x}\right) + c \). center Final Answer: (A) -12 ^-1(x+2x) + c center
Step 1: Simplify the expression inside the square root.
Factor out \(x^2\) from the square root: \[ \sqrt{x^4 + 2x^3 + 2x^2} = \sqrt{x^2(x^2 + 2x + 2)} = x\sqrt{x^2 + 2x + 2} \] So the integral becomes: \[ I = \int \frac{1}{x\sqrt{x^2 + 2x + 2}} \, dx \]
Step 2: Apply substitution to simplify the integrand.
Let \(x = \frac{1}{t}\), then \(dx = -\frac{1}{t^2} dt\). Substituting these into the integral: \[ I = \int \frac{1}{(1/t)\sqrt{(1/t)^2 + 2(1/t) + 2}} \left(-\frac{1}{t^2}\right) \, dt \] \[ I = -\int \frac{1}{t\sqrt{\frac{1 + 2t + 2t^2}{t^2}}} \, dt = -\int \frac{1}{\sqrt{2t^2 + 2t + 1}} \, dt \]
Step 3: Complete the square and integrate.
Factor out \(\sqrt{2}\) from the square root: \[ I = -\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{t^2 + t + 1/2}} \, dt \] Completing the square for \(t^2 + t + 1/2 = (t + 1/2)^2 + 1/4\): \[ I = -\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{(t + 1/2)^2 + (1/2)^2}} \, dt \] Using the standard integral \(\int \frac{1}{\sqrt{u^2 + a^2}} du = \sinh^{-1}(u/a)\): \[ I = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{t + 1/2}{1/2}\right) + c = -\frac{1}{\sqrt{2}} \sinh^{-1}(2t + 1) + c \] Since \(t = 1/x\): \[ I = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{2}{x} + 1\right) + c = -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x + 2}{x}\right) + c \]
Conclusion: The value of the integral is \( -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x+2}{x}\right) + c \). center Final Answer: (A) -12 ^-1(x+2x) + c center
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