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9. Evaluate the indefinite integral. (Use \( C \) for the constant of integration.) \[ \int \frac{a+b x^{5}}{\sqrt{6 a x+b x^{6}}} d x \]

Ask by Boyd Paul. in Bahrain
Feb 20,2025

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The integral \( \int \frac{a+b x^{5}}{\sqrt{6 a x+b x^{6}}} d x \) equals \( \frac{1}{3}\sqrt{6ax+bx^{6}} + C \).

Solución

Calculate the integral \( \int \frac{a+b x^{5}}{\sqrt{6 a x+b x^{6}}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int \frac{a+bx^{5}}{\sqrt{6ax+bx^{6}}} dx\) - step1: Rewrite the expression: \(\int \frac{a+bx^{5}}{\left(6ax+bx^{6}\right)^{\frac{1}{2}}} dx\) - step2: Use the substitution \(dx=\frac{1}{6a+6bx^{5}} dt\) to transform the integral\(:\) \(\int \frac{a+bx^{5}}{\left(6ax+bx^{6}\right)^{\frac{1}{2}}}\times \frac{1}{6a+6bx^{5}} dt\) - step3: Simplify: \(\int \frac{1}{6\left(6ax+bx^{6}\right)^{\frac{1}{2}}} dt\) - step4: Use the substitution \(t=6ax+bx^{6}\) to transform the integral\(:\) \(\int \frac{1}{6t^{\frac{1}{2}}} dt\) - step5: Rewrite the expression: \(\int \frac{1}{6}\times \frac{1}{t^{\frac{1}{2}}} dt\) - step6: Use properties of integrals: \(\frac{1}{6}\times \int \frac{1}{t^{\frac{1}{2}}} dt\) - step7: Evaluate the integral: \(\frac{1}{6}\times \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) - step8: Simplify: \(\frac{1}{6}\times 2t^{\frac{1}{2}}\) - step9: Multiply the numbers: \(\frac{1}{3}t^{\frac{1}{2}}\) - step10: Substitute back: \(\frac{1}{3}\left(6ax+bx^{6}\right)^{\frac{1}{2}}\) - step11: Simplify: \(\frac{1}{3}\sqrt{6ax+bx^{6}}\) - step12: Add the constant of integral C: \(\frac{1}{3}\sqrt{6ax+bx^{6}} + C, C \in \mathbb{R}\) The indefinite integral of \( \frac{a+b x^{5}}{\sqrt{6 a x+b x^{6}}} \) with respect to \( x \) is \( \frac{1}{3}\sqrt{6ax+bx^{6}} + C \), where \( C \) is the constant of integration.

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To evaluate the indefinite integral \[ \int \frac{a + b x^{5}}{\sqrt{6 a x + b x^{6}}} \, dx, \] let's use a substitution. Set \[ u = 6 a x + b x^{6}. \] Then compute the derivative of \( u \): \[ \frac{du}{dx} = 6a + 6bx^{5}, \] which simplifies to \[ du = (6a + 6bx^{5}) dx = 6(a + bx^{5}) \, dx. \] Now we can express \( dx \): \[ dx = \frac{du}{6(a + bx^{5})}. \] Substituting in our integral, we have: \[ \int \frac{a + b x^{5}}{\sqrt{u}} \cdot \frac{du}{6(a + bx^{5})}. \] Notice that the \( (a + bx^{5}) \) terms cancel, leading to: \[ \frac{1}{6} \int \frac{1}{\sqrt{u}} \, du. \] Now, we can easily integrate this: \[ \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C. \] Substituting back \( u = 6 a x + b x^{6} \), we find: \[ \frac{1}{6} \cdot 2\sqrt{6 a x + b x^{6}} + C = \frac{1}{3}\sqrt{6 a x + b x^{6}} + C. \] Thus, the final result for the indefinite integral is: \[ \int \frac{a + b x^{5}}{\sqrt{6 a x + b x^{6}}} \, dx = \frac{1}{3}\sqrt{6 a x + b x^{6}} + C. \]

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