Let's solve the integral step-by-step using each substitution. The integral is: \[ \int \frac{x^3}{\sqrt{x^2 + 1}}\, dx \] --- ### **SUBSTITUTION 1: \( u = x^2 + 1 \)** #### **Transformed Integral:** Let \( u = x^2 + 1 \implies du = 2x\,dx \implies x\,dx = \frac{du}{2} \). But in the integral, we have \( x^3\, dx \): \[ x^3\, dx = x^2 \cdot x\, dx = (u-1)\frac{du}{2} \] Now, substitute \( x^2 = u-1 \) and \( \sqrt{x^2+1} = \sqrt{u} \): \[ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \int \frac{(u-1)x}{\sqrt{u}} \cdot \frac{du}{2} \] But recall \( x = \sqrt{u-1} \): \[ = \int \frac{(u-1)\sqrt{u-1}}{\sqrt{u}} \cdot \frac{du}{2} \] Simplify \(\frac{(u-1)\sqrt{u-1}}{\sqrt{u}} = (u-1) \cdot \sqrt{\frac{u-1}{u}} \): \[ = \frac{1}{2} \int (u-1) \sqrt{\frac{u-1}{u}}\, du \] #### **Details:** - \( u = x^2 + 1 \) - \( du = 2x\, dx \implies x\, dx = \frac{du}{2} \) - \( x^2 = u-1 \), \( x = \sqrt{u-1} \) - Substitution leads to: \[ \boxed{ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \frac{1}{2} \int (u-1) \sqrt{\frac{u-1}{u}}\, du } \] --- ### **SUBSTITUTION 2: \( v = \sqrt{x^2 + 1} \)** #### **Transformed Integral:** Let \( v = \sqrt{x^2+1} \implies v^2 = x^2+1 \implies x^2 = v^2-1 \). \[ 2v\, dv = 2x\, dx \implies v\, dv = x\, dx \implies dx = \frac{v\, dv}{x} \] Now, \( x = \sqrt{v^2-1} \): \[ \int \frac{x^3}{v}\, dx = \int \frac{(\sqrt{v^2-1})^3}{v} \cdot \frac{v}{\sqrt{v^2-1}}\, dv \] Now, \((\sqrt{v^2-1})^3 = (v^2-1)^{3/2}\) and \( \frac{v}{\sqrt{v^2-1}} \) is from \( dx \): \[ = \int \frac{(v^2-1)^{3/2}}{v} \cdot \frac{v}{\sqrt{v^2-1}}\, dv \] \[ = \int (v^2-1)^{3/2} \cdot \frac{1}{\sqrt{v^2-1}}\, dv \] \[ = \int (v^2-1) \, dv \] #### **Details:** - \( v = \sqrt{x^2+1} \) - \( v^2 = x^2 + 1 \implies x^2 = v^2-1 \), \( x = \sqrt{v^2-1} \) - \( dx = \frac{v\, dv}{x} \) - Substitution leads to: \[ \boxed{ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \int (v^2-1) \, dv } \] --- ### **SUBSTITUTION 3: \( x = \tan\theta \)** #### **Transformed Integral:** Let \( x = \tan\theta \implies dx = \sec^2\theta\, d\theta \), \( x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta \), \( \sqrt{x^2+1} = \sec\theta \): \[ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \int \frac{\tan^3\theta}{\sec\theta} \cdot \sec^2\theta\, d\theta \] \[ = \int \tan^3\theta \cdot \sec\theta\, d\theta \] #### **Details:** - \( x = \tan\theta \) - \( dx = \sec^2\theta\, d\theta \) - \( \sqrt{x^2+1} = \sec\theta \) - Substitution leads to: \[ \boxed{ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \int \tan^3\theta \sec\theta\, d\theta } \] --- ## **Final Antiderivative (using Substitution 2 for simplicity):** From Substitution 2, \[ \int (v^2-1) dv = \int v^2\, dv - \int dv = \frac{v^3}{3} - v + C \] Recall \( v = \sqrt{x^2+1} \): \[ \boxed{ \int \frac{x^3}{\sqrt{x^2+1}}\, dx = \frac{(x^2+1)^{3/2}}{3} - (x^2+1)^{1/2} + C } \]

The integral \(\int \frac{x^3}{\sqrt{x^2 + 1}}\, dx\) can be solved using different substitutions. The most straightforward one is substitution 2: **Using substitution \( v = \sqrt{x^2 + 1} \):** - The transformed integral becomes: \[ \int (v^2 - 1) \, dv \] - Integrating: \[ \frac{v^3}{3} - v + C \] - Re-substituting \( v = \sqrt{x^2 + 1} \): **Final answer:** \[ \boxed{ \int \frac{x^3}{\sqrt{x^2 + 1}}\, dx = \frac{(x^2 + 1)^{3/2}}{3} - \sqrt{x^2 + 1} + C } \]