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Evaluate the integral. (Use \( C \) for the constant of integration.) \[ \int x^{3} \sqrt{9+x^{2}} d x \] \( \square \)

Ask by Pope Paul. in the United States
Feb 25,2025

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Answer

The integral of \( x^{3} \sqrt{9+x^{2}} \) with respect to \( x \) is: \[ \frac{(9+x^{2})^{2}\sqrt{9+x^{2}}}{5} - 27\sqrt{9+x^{2}} - 3x^{2}\sqrt{9+x^{2}} + C \] where \( C \) is the constant of integration.

Solution

Calculate the integral \( \int x^{3} \sqrt{9+x^{2}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int x^{3}\sqrt{9+x^{2}} dx\) - step1: Rewrite the expression: \(\int x^{3}\left(9+x^{2}\right)^{\frac{1}{2}} dx\) - step2: Use the substitution \(dx=\frac{1}{2x} dt\) to transform the integral\(:\) \(\int x^{3}\left(9+x^{2}\right)^{\frac{1}{2}}\times \frac{1}{2x} dt\) - step3: Simplify: \(\int \frac{\left(9+x^{2}\right)^{\frac{1}{2}}x^{2}}{2} dt\) - step4: Use the substitution \(t=x^{2}\) to transform the integral\(:\) \(\int \frac{\left(9+t\right)^{\frac{1}{2}}t}{2} dt\) - step5: Rewrite the expression: \(\int \frac{1}{2}\left(9+t\right)^{\frac{1}{2}}t dt\) - step6: Use properties of integrals: \(\frac{1}{2}\times \int \left(9+t\right)^{\frac{1}{2}}t dt\) - step7: Use the substitution \(dt=1 dv\) to transform the integral\(:\) \(\frac{1}{2}\times \int \left(9+t\right)^{\frac{1}{2}}t\times 1 dv\) - step8: Simplify: \(\frac{1}{2}\times \int \left(9+t\right)^{\frac{1}{2}}t dv\) - step9: Use the substitution \(v=9+t\) to transform the integral\(:\) \(\frac{1}{2}\times \int v^{\frac{1}{2}}\left(v-9\right) dv\) - step10: Multiply the terms: \(\frac{1}{2}\times \int \left(v^{\frac{3}{2}}-9v^{\frac{1}{2}}\right) dv\) - step11: Use properties of integrals: \(\frac{1}{2}\left(\int v^{\frac{3}{2}} dv+\int -9v^{\frac{1}{2}} dv\right)\) - step12: Calculate: \(\frac{1}{2}\times \int v^{\frac{3}{2}} dv+\frac{1}{2}\times \int -9v^{\frac{1}{2}} dv\) - step13: Evaluate the integral: \(\frac{v^{\frac{5}{2}}}{5}+\frac{1}{2}\times \int -9v^{\frac{1}{2}} dv\) - step14: Evaluate the integral: \(\frac{v^{\frac{5}{2}}}{5}-3v^{\frac{3}{2}}\) - step15: Substitute back: \(\frac{\left(9+t\right)^{\frac{5}{2}}}{5}-3\left(9+t\right)^{\frac{3}{2}}\) - step16: Substitute back: \(\frac{\left(9+x^{2}\right)^{\frac{5}{2}}}{5}-3\left(9+x^{2}\right)^{\frac{3}{2}}\) - step17: Simplify: \(\frac{\sqrt{\left(9+x^{2}\right)^{5}}}{5}-3\left(9+x^{2}\right)^{\frac{3}{2}}\) - step18: Simplify: \(\frac{\sqrt{\left(9+x^{2}\right)^{5}}}{5}-3\sqrt{\left(9+x^{2}\right)^{3}}\) - step19: Expand the expression: \(\frac{\left(9+x^{2}\right)^{2}\sqrt{9+x^{2}}}{5}+\left(-27-3x^{2}\right)\sqrt{9+x^{2}}\) - step20: Expand the expression: \(\frac{\left(9+x^{2}\right)^{2}\sqrt{9+x^{2}}}{5}-27\sqrt{9+x^{2}}-3x^{2}\sqrt{9+x^{2}}\) - step21: Add the constant of integral C: \(\frac{\left(9+x^{2}\right)^{2}\sqrt{9+x^{2}}}{5}-27\sqrt{9+x^{2}}-3x^{2}\sqrt{9+x^{2}} + C, C \in \mathbb{R}\) The integral of \(x^{3} \sqrt{9+x^{2}}\) with respect to \(x\) is given by: \[ \int x^{3} \sqrt{9+x^{2}} d x = \frac{(9+x^{2})^{2}\sqrt{9+x^{2}}}{5} - 27\sqrt{9+x^{2}} - 3x^{2}\sqrt{9+x^{2}} + C \] where \(C\) is the constant of integration.

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To evaluate the integral \[ \int x^{3} \sqrt{9+x^{2}} \, dx, \] we can use the substitution \( u = 9 + x^2 \). Then, \( du = 2x \, dx \) or \( \frac{du}{2} = x \, dx \). We also notice that \( x^2 = u - 9 \), so \( x^3 = x(x^2) = x(u - 9) \). Expressing \( x \) in terms of \( u \) gives us \( x = \sqrt{u - 9} \), hence \( x^3 = \sqrt{u - 9}(u - 9) = (u - 9)^{1/2}(u - 9) = (u - 9)^{3/2}\). The integral becomes: \[ \int (u - 9)^{3/2} \cdot \frac{du}{2\sqrt{u - 9}} = \frac{1}{2} \int (u - 9) \sqrt{u} \, du. \] This can be expanded and simplified. We can change back to \( u \) and find the antiderivative: \[ \frac{1}{2} \int (u^{3/2} - 9u^{1/2}) \, du = \frac{1}{2} \left(\frac{2}{5}u^{5/2} - 6u^{3/2} \right) + C. \] Finally, substituting back \( u = 9 + x^2 \): \[ \frac{1}{5}(9+x^2)^{5/2} - 3(9+x^2)^{3/2} + C. \] Thus, the result of the integral is: \[ \frac{1}{5}(9+x^2)^{5/2} - 3(9+x^2)^{3/2} + C. \] Have fun integrating and remember: practice makes perfect!

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