Evaluate the integral: I = integral (5th root of (1 - 2x

Evaluate the integral: $$ I = \int \frac{\sqrt[5]{1 - 2x + x^2}}{1 - x} \, dx. $$

Step 1 · Simplify the expression inside the radical

Notice that the expression inside the fifth root, $1-2x+x^2$, is a perfect square:

$$ 1 - 2x + x^2 = (1-x)^2. $$

Thus, the integral becomes:

$$ I = \int \frac{\sqrt[5]{(1-x)^2}}{1-x}\,dx. $$

Step 2 · Simplify the integrand

Since $\sqrt[5]{(1-x)^2} = (1-x)^{2/5}$, the integral now becomes:

$$ I = \int \frac{(1-x)^{2/5}}{1-x}\,dx. $$

Now, simplify the expression $\dfrac{(1-x)^{2/5}}{1-x}$ as:

$$ \frac{(1-x)^{2/5}}{1-x} = (1-x)^{2/5-1} = (1-x)^{-3/5}. $$

Thus, the integral becomes:

$$ I = \int (1-x)^{-3/5}\,dx. $$

Step 3 · Use substitution

Let us use substitution to simplify the integral. Let:

$$ u = 1 - x \quad \Rightarrow \quad du = -dx. $$

Substituting this into the integral, we get:

$$ I = \int u^{-3/5}(-du) = -\int u^{-3/5}\,du. $$

Step 4 · Integrate using the power rule

We now apply the power rule of integration:

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

Here, $n = -3/5$, so we have:

$$ \int u^{-3/5}\,du = \frac{u^{-3/5+1}}{-3/5+1} = \frac{u^{2/5}}{2/5} = \frac{5}{2}u^{2/5}. $$

Step 5 · Substitute back $u = 1 - x$

Substituting back $u = 1-x$ into the result, we get:

$$ I = -\frac{5}{2}(1-x)^{2/5} + C. $$

Summary Table

Additional Notes

• The domain of $I$ requires $x \neq 1$, since $1-x$ appears in a denominator before simplification (and the exponent $-3/5$ after simplification is still undefined at $x=1$).
• Spotting the perfect square $1-2x+x^2=(1-x)^2$ before touching the radical is the key simplification — without it, the fifth root and the denominator would need to be handled together, which is messier.
• The result can be checked by differentiating $-\tfrac52(1-x)^{2/5}$ with the chain rule and confirming it returns $(1-x)^{-3/5}$.