Evaluate the integral: I = integral (x + 5)/(5th root
Evaluate the integral: $$ I = \int \frac{x + 5}{\sqrt[5]{1 - 2x}} \, dx. $$
Explanation of Method
With a linear numerator and a root of a different linear expression in the denominator, substituting $u$ for the expression inside the root and solving for $x$ in terms of $u$ turns the whole integrand into a difference of two power terms divided by $u^{1/5}$, ready for the power rule.
- Substitute $u$ for the expression inside the root.
- Rewrite the numerator entirely in terms of $u$.
- Split the integral and integrate each power term.
- Substitute back and add the constant of integration.
Step 1 · Use substitution
To simplify this integral, we use substitution. Let:
$$ u = 1 - 2x. $$
Then, the derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = -2 \quad \Rightarrow \quad du = -2\,dx. $$
Solving for $dx$, we get:
$$ dx = \frac{du}{-2}. $$
Now, express $x$ in terms of $u$:
$$ x = \frac{1-u}{2}. $$
Substitute into the integral:
$$ I = \int \frac{\frac{1-u}{2}+5}{\sqrt[5]{u}} \cdot \frac{du}{-2}. $$
Simplifying the numerator:
$$ \frac{1-u}{2}+5 = \frac{1-u+10}{2} = \frac{11-u}{2}. $$
So, the integral becomes:
$$ I = -\frac{1}{4}\int \frac{11-u}{u^{1/5}}\,du. $$
Step 2 · Break into two integrals
Now break the integral into two separate integrals:
$$ I = -\frac{1}{4}\left(\int \frac{11}{u^{1/5}}\,du - \int \frac{u}{u^{1/5}}\,du\right). $$
Simplify each term:
$$ I = -\frac{1}{4}\left(11\int u^{-1/5}\,du - \int u^{4/5}\,du\right). $$
Step 3 · Integrate each term
First term:
$$ \int u^{-1/5}\,du = \frac{u^{4/5}}{4/5} = \frac{5}{4}u^{4/5}. $$
Multiply by $11$:
$$ 11\int u^{-1/5}\,du = 11\cdot \frac{5}{4}u^{4/5} = \frac{55}{4}u^{4/5}. $$
Second term:
$$ \int u^{4/5}\,du = \frac{u^{9/5}}{9/5} = \frac{5}{9}u^{9/5}. $$
Step 4 · Substitute back $u = 1 - 2x$
Substitute $u = 1-2x$ back into the results:
$$ I = -\frac{1}{4}\left(\frac{55}{4}(1-2x)^{4/5} - \frac{5}{9}(1-2x)^{9/5}\right). $$
Step 5 · Final simplification
Now, distribute the constants:
$$ I = -\frac{55}{16}(1-2x)^{4/5} + \frac{5}{36}(1-2x)^{9/5} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $-\tfrac14\cdot 11 u^{-1/5}$ | $-\tfrac{55}{16}(1-2x)^{4/5}$ |
| $-\tfrac14\cdot(-u^{4/5})$ | $\tfrac{5}{36}(1-2x)^{9/5}$ |
Additional Notes
- The integrand is defined for all real $x$, since the fifth root (an odd root) is defined for both positive and negative arguments, including $u=0$ needing care only if it appears in a denominator with a negative exponent — here $u^{-1/5}$ requires $u\neq 0$, i.e. $x\neq \tfrac12$.
- Carrying the constant factor $-\tfrac14$ through both terms of the split integral, rather than recomputing it separately for each, keeps the bookkeeping simpler.
- The result can be checked by differentiating $-\tfrac{55}{16}(1-2x)^{4/5}+\tfrac{5}{36}(1-2x)^{9/5}$ with the chain rule and confirming it returns $\dfrac{x+5}{\sqrt[5]{1-2x}}$.
Result
The solution to the integral is: