Evaluate the integral: I = integral x 8th root of (4+3x)
Evaluate the integral: $$ I = \int x\sqrt[8]{4+3x}\,dx. $$
Explanation of Method
Substituting $u=4+3x$ turns the fractional power into a simple power of $u$, and $x$ into a linear function of $u$, reducing the integral to two applications of the power rule.
- Substitute $u=4+3x$.
- Expand the integrand as powers of $u$.
- Integrate term by term.
- Back-substitute and simplify.
Step 1 · Substitute
Let $u=4+3x$, so $x=\dfrac{u-4}3$ and $dx=\dfrac{du}3$:
$$ I = \int \frac{u-4}3\cdot u^{1/8}\cdot\frac{du}3 = \frac19\int\left(u^{9/8}-4u^{1/8}\right)du. $$
Step 2 · Integrate term by term
$$ \frac19\left[\frac{8u^{17/8}}{17} - 4\cdot\frac{8u^{9/8}}9\right] + C = \frac{8u^{17/8}}{153} - \frac{32u^{9/8}}{81} + C. $$
Step 3 · Back-substitute and combine
Factoring out $u^{9/8}$ and combining over the common denominator $1377=9\cdot153=17\cdot81$:
$$ I = u^{9/8}\left[\frac{8u}{153}-\frac{32}{81}\right]+C = \frac{u^{9/8}\left[72u-544\right]}{1377}+C = \frac{8u^{9/8}(9u-68)}{1377}+C. $$
Substituting $u=4+3x$, so $9u-68=9(4+3x)-68=27x-32$:
$$ I = \frac{8(4+3x)^{9/8}(27x-32)}{1377}+C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u^{9/8}$ | $\tfrac89u^{17/8}$ |
| $-4u^{1/8}$ | $-\tfrac{32}9u^{9/8}$ |
Additional Notes
- The domain requires $4+3x\geq0$, i.e. $x\geq-\tfrac43$, for the eighth root to be real.
- The result was verified by direct expansion: reversing the substitution and differentiating $\tfrac{8}{153}(4+3x)^{17/8}-\tfrac{32}{81}(4+3x)^{9/8}$ reproduces $x(4+3x)^{1/8}$ after simplification.
Result
The solution to the integral is: