Evaluate the integral (for x ≥0): I = integral

Evaluate the integral (for $x\geq0$): $$ I = \int \sqrt{x^3+x^4}\,dx = \int x^{3/2}\sqrt{1+x}\,dx. $$

Step 1 · Substitute

Let $x=u^2$ ($u\geq0$), so $dx=2u\,du$, $x^{3/2}=u^3$, $1+x=1+u^2$:

$$ I = \int u^3\sqrt{1+u^2}\cdot2u\,du = 2\int u^4\sqrt{1+u^2}\,du. $$

Step 2 · Apply the reduction

Using the standard reduction $\int u^4\sqrt{u^2+1}\,du=\tfrac{u^3(u^2+1)^{3/2}}6-\tfrac18\left[u(u^2+1)^{3/2}-\int u\sqrt{u^2+1}^{}\,du\cdot3\right]$-style bookkeeping (equivalently, integrating by parts twice and combining), the result is:

$$ \int u^4\sqrt{u^2+1}\,du = \frac{u\left(8u^6+10u^4-u^2-3\right)}{48\sqrt{u^2+1}} + \frac1{16}\,\text{arcsinh}(u). $$

Rationalizing the first term over $\sqrt{u^2+1}$ and doubling:

$$ I = \frac{u\left(8u^6+10u^4-u^2-3\right)\sqrt{u^2+1}}{24(u^2+1)} + \frac18\,\text{arcsinh}(u) + C. $$

Since $u^2+1$ divides evenly (the numerator factors as $(u^2+1)(8u^4+2u^2-3)$), this simplifies to:

$$ I = \frac{u\left(8u^4+2u^2-3\right)\sqrt{u^2+1}}{24} + \frac18\,\text{arcsinh}(u) + C. $$

Step 3 · Back-substitute

With $u=\sqrt x$, $u^2=x$, $u^4=x^2$, $\sqrt{u^2+1}=\sqrt{1+x}$:

Summary Table

Additional Notes

• Major rewrite: the source's Step 5–6 abandoned two different substitution attempts as "too complex" and then asserted, without derivation, two mutually different boxed answers ($\tfrac{2x^{3/2}(1+x)^{3/2}}5-\tfrac{2x^{5/2}\sqrt{1+x}}7$ and $\tfrac{2\sqrt{x^3+x^4}(7+2x)}{35}$). Testing the second at $x=1$ gives a derivative of $\approx1.434$, while the true integrand there is $\approx1.414$ — close but not exact, confirming it as an unverified guess. The boxed result below was obtained via the substitution $x=u^2$ (which converts the half-integer powers into whole powers) followed by the standard reduction formula, and was verified independently.
• The domain is $x\geq0$.
• The result was verified by differentiation at $x=1,2,4$: the derivative matches the integrand $\sqrt{x^3+x^4}$ exactly at all three points.