Evaluate the integral: I = integral (x²)/((1-x)^100) dx
Evaluate the integral: $$ I = \int \frac{x^2}{(1-x)^{100}}\,dx. $$
Explanation of Method
Substituting $u=1-x$ turns the numerator $x^2$ into a simple binomial $(1-u)^2$ in $u$, while the denominator becomes a plain power $u^{100}$. Expanding the numerator then reduces the whole integral to three power-rule terms.
- Substitute $u=1-x$.
- Expand $(1-u)^2$ and split into three terms.
- Integrate each term using the power rule.
- Substitute back and add the constant of integration.
Step 1 · Substitution
Let:
$$ u = 1-x \quad \text{so that} \quad x = 1-u, \quad du = -dx. $$
Substituting into the integral:
$$ I = \int \frac{(1-u)^2}{u^{100}}(-du). $$
Step 2 · Expand and split
Expanding $(1-u)^2=1-2u+u^2$:
$$ I = -\int \left(u^{-100} - 2u^{-99} + u^{-98}\right) du. $$
Step 3 · Integrate each term
Using the power rule on each term:
$$ \int u^{-100}\,du = -\frac{1}{99u^{99}}, \qquad \int u^{-99}\,du = -\frac{1}{98u^{98}}, \qquad \int u^{-98}\,du = -\frac{1}{97u^{97}}. $$
Combining with the overall negative sign:
$$ I = \frac{1}{99u^{99}} - \frac{2}{98u^{98}} + \frac{1}{97u^{97}} + C = \frac{1}{99u^{99}} - \frac{1}{49u^{98}} + \frac{1}{97u^{97}} + C. $$
Step 4 · Substitute back
Substituting back $u=1-x$:
$$ I = \frac{1}{99(1-x)^{99}} - \frac{1}{49(1-x)^{98}} + \frac{1}{97(1-x)^{97}} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $-u^{-100}$ | $1/(99u^{99})$ |
| $2u^{-99}$ | $-1/(49u^{98})$ |
| $-u^{-98}$ | $1/(97u^{97})$ |
Additional Notes
- The domain excludes $x=1$, where the original denominator vanishes.
- This is the same idea as Problem 114's polynomial division, but pushed to a much higher power: substituting $u$ for the linear binomial and expanding the (here, low-degree) numerator avoids ever having to expand $(1-x)^{100}$ directly.
- The result can be checked by differentiating each term and confirming the sum returns $x^2/(1-x)^{100}$, using $x=1-u$ throughout.
Result
The solution to the integral is: