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int(dx)/(sqrt(25+4x^(2)))

∫dx25+4x2 \int \frac{d x}{\sqrt{25+4 x^{2}}}

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Find indefinite integrals using the substitution and by partsright chevron icon

Full solution

Q. ∫dx25+4x2 \int \frac{d x}{\sqrt{25+4 x^{2}}}
  1. Rewrite with Substitution: Rewrite the integral with a substitution to simplify the square root term. Let u=2xu = 2x, then du=2dxdu = 2dx, or dx=du2dx = \frac{du}{2}.
  2. Substitute and Simplify: Substitute into the integral: ∫dx25+4x2=∫du2125+u2\int \frac{dx}{\sqrt{25+4x^{2}}} = \int \frac{du}{2}\frac{1}{\sqrt{25+u^2}}.
  3. Recognize Standard Integral: Simplify the integral: ∫du2/25+u2=12∫du25+u2\int \frac{du}{2} / \sqrt{25 + u^2} = \frac{1}{2} \int \frac{du}{\sqrt{25 + u^2}}.
  4. Calculate Integral: Recognize the integral form: ∫dua2+u2\int \frac{du}{\sqrt{a^2+u^2}} is a standard integral, which equals arcsinh(ua)+C\text{arcsinh}(\frac{u}{a}) + C, where a=5a = 5 in this case.
  5. Substitute Back: Calculate the integral: 12∫du25+u2=12arcsinh(u5)+C\frac{1}{2} \int \frac{du}{\sqrt{25+u^2}} = \frac{1}{2} \text{arcsinh}\left(\frac{u}{5}\right) + C.
  6. Substitute Back: Calculate the integral: 12∫du25+u2=12arcsinh(u5)+C\frac{1}{2} \int \frac{du}{\sqrt{25+u^2}} = \frac{1}{2} \text{arcsinh}\left(\frac{u}{5}\right) + C. Substitute back for uu: 12arcsinh(u5)+C=12arcsinh(2x5)+C\frac{1}{2} \text{arcsinh}\left(\frac{u}{5}\right) + C = \frac{1}{2} \text{arcsinh}\left(\frac{2x}{5}\right) + C.

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