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kondaur [170]
2 years ago
7

Evaluate the indefinite integral as a power series. t 1 − t7 dt

Mathematics
1 answer:
sammy [17]2 years ago
4 0
Given

\int\limits { \frac{t}{1-t^7} } \, dt

Recall that the general geometric series representation is given by

\Sigma_0^\infty x^k= \frac{1}{1-x}

Let x=t^7

Thus,

\frac{1}{1-t^7}=\Sigma_0^\infty (t^7)^k=\Sigma_0^\infty t^{7k}

Multiply both sides by t, to get

\frac{1}{1-t^7}=\Sigma_0^\infty t^{7k+1}

Integrate both sides to get

\int { \frac{t}{1-t^7} } \, dt=\Sigma_0^\infty  \frac{t^{7k+2}}{7k+2}
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