Evaluate the indefinite integral as a power series: ∫11−t9 dt.

∑ n = 0 ∞ t 9 n + 1 9 n + 1 + C

Evaluate the indefinite integral as a power series: ∫ 1 1 − t 9 d t .

∑ n = 0 ∞ t 9 n + 1 9 n + 1 + C

∑ n = 0 ∞ t n + 1 n + 1 + C

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
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10. Physics Applications of Integrals 3h 16m
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  1h 13m
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11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
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- Taylor Series & Taylor Polynomials
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16. Parametric Equations & Polar Coordinates7h 58m

7. Antiderivatives & Indefinite Integrals

Indefinite Integrals

Calculus

7. Antiderivatives & Indefinite Integrals

Indefinite Integrals

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Multiple Choice

Evaluate the indefinite integral as a power series: $\int \frac{1}{1 - t^{9}} d t$ .

\sum n = 1 \infty \frac{t^{n}}{n}

+ C

\sum n = 0 \infty t^{9 n}

+ C

\sum n = 0 \infty \frac{t^{n + 1}}{n + 1}

+ C

\sum n = 0 \infty \frac{t^{9 n + 1}}{9 n + 1}

+ C

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Verified step by step guidance

Recognize that the given integral $ \int \frac{1}{1 - t^9} \, dt $ involves a geometric series expansion. Recall that the geometric series $ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n $ for $ |x| < 1 $. Here, $ x = t^9 $.

Substitute $ t^9 $ for $ x $ in the geometric series expansion. This gives $ \frac{1}{1 - t^9} = \sum_{n=0}^{\infty} t^{9n} $.

Integrate term by term. The integral of $ t^{9n} $ is $ \frac{t^{9n+1}}{9n+1} $, assuming $ 9n+1 \neq 0 $. Add the constant of integration $ C $ at the end.

Write the resulting power series after integration: $ \int \frac{1}{1 - t^9} \, dt = \sum_{n=0}^{\infty} \frac{t^{9n+1}}{9n+1} + C $.

Verify the result by differentiating the power series $ \sum_{n=0}^{\infty} \frac{t^{9n+1}}{9n+1} $. Differentiating term by term should yield $ \frac{1}{1 - t^9} $, confirming the correctness of the solution.