Question

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

Answer 1

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

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