Evaluate the integral: I = integral x³ cube root of (1 +

Evaluate the integral: $$ I = \int x^3 \sqrt[3]{1 + x^2} \, dx. $$

Step 1 · Substitution

Let's perform the substitution:

$$ u = 1 + x^2. $$

Then, differentiate both sides with respect to $x$:

$$ du = 2x\,dx. $$

Now we need to express $x^3\,dx$ in terms of $du$. Notice that:

$$ x^3\,dx = x\cdot x^2\,dx = x\cdot(u-1)\,dx. $$

Since $x\,dx = \dfrac{du}{2}$, we substitute this into the equation:

$$ x^3\,dx = \frac{(u-1)}{2}\,du. $$

Step 2 · Substitute into the integral

Now, substitute the expression for $x^3\,dx$ and $u=1+x^2$ into the integral:

$$ \int x^3\sqrt[3]{1+x^2}\,dx = \int \frac{(u-1)}{2}\cdot u^{1/3}\,du. $$

Simplify the expression:

$$ = \frac{1}{2}\int (u-1)u^{1/3}\,du. $$

Distribute $u^{1/3}$:

$$ = \frac{1}{2}\int \left(u^{4/3} - u^{1/3}\right) du. $$

Step 3 · Integrate term by term

Now we can integrate each term separately:

$$ \int u^{4/3}\,du = \frac{u^{7/3}}{7/3} = \frac{3u^{7/3}}{7}, \qquad \int u^{1/3}\,du = \frac{u^{4/3}}{4/3} = \frac{3u^{4/3}}{4}. $$

Thus, the integral becomes:

$$ \frac{1}{2}\left(\frac{3u^{7/3}}{7} - \frac{3u^{4/3}}{4}\right) = \frac{3u^{7/3}}{14} - \frac{3u^{4/3}}{8}. $$

Step 4 · Substitute back $u = 1 + x^2$

Finally, substitute $u = 1+x^2$ back into the solution:

$$ I = \frac{3\left(1+x^2\right)^{7/3}}{14} - \frac{3\left(1+x^2\right)^{4/3}}{8} + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $1+x^2>0$ always and cube roots are defined for any real number.
• Whenever the power of $x$ outside a root is more than one degree higher than what the derivative of the inside expression provides, expressing the leftover factor(s) in terms of $u$ (here, $x^2=u-1$) before integrating is the standard technique.
• The result can be checked by differentiating $\tfrac{3}{14}(1+x^2)^{7/3} - \tfrac38(1+x^2)^{4/3}$ with the chain rule and confirming it returns $x^3\sqrt[3]{1+x^2}$.