Evaluate the integral: I = integral 5th root of (1 - x) dx

Evaluate the integral: $$ I = \int \sqrt[5]{1 - x} \, dx. $$

Step 1 · Use substitution

To simplify the integral, we will use substitution. Let:

$$ u = 1 - x. $$

Then, differentiate $u$ with respect to $x$:

$$ \frac{du}{dx} = -1 \quad \Rightarrow \quad dx = -du. $$

Substituting into the integral, we get:

$$ I = \int \sqrt[5]{u}\,(-du) = -\int u^{1/5}\,du. $$

Step 2 · Integrate

We now integrate $u^{1/5}$. Using the power rule for integration:

$$ \int u^n\,du = \frac{u^{n+1}}{n+1} + C \quad \text{for } n\neq -1, $$

we get:

$$ \int u^{1/5}\,du = \frac{u^{6/5}}{6/5} = \frac{5}{6}u^{6/5}. $$

Step 3 · Substitute back $u = 1 - x$

Now, substitute back $u = 1-x$ into the result:

$$ I = -\frac{5}{6}(1-x)^{6/5}. $$

Step 4 · Add the constant of integration

Finally, we add the constant of integration $C$ to the result:

$$ I = -\frac{5}{6}(1-x)^{6/5} + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since odd roots (like the fifth root) are defined for negative arguments as well as positive ones.
• The negative sign in $dx=-du$ carries through the entire computation, flipping the sign of the final antiderivative relative to what a plain power rule application on $u$ would give.
• The result can be checked by differentiating $-\tfrac{5}{6}(1-x)^{6/5}$ with the chain rule and confirming it returns $(1-x)^{1/5} = \sqrt[5]{1-x}$.