johnkhan5526 johnkhan5526
  • 21-11-2017
  • Mathematics
contestada

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

Respuesta :

nobillionaireNobley
nobillionaireNobley nobillionaireNobley
  • 01-12-2017
Given

[tex] \int\limits { \frac{t}{1-t^7} } \, dt [/tex]

Recall that the general geometric series representation is given by

[tex]\Sigma_0^\infty x^k= \frac{1}{1-x} [/tex]

Let [tex]x=t^7[/tex]

Thus,

[tex]\frac{1}{1-t^7}=\Sigma_0^\infty (t^7)^k=\Sigma_0^\infty t^{7k}[/tex]

Multiply both sides by t, to get

[tex]\frac{1}{1-t^7}=\Sigma_0^\infty t^{7k+1}[/tex]

Integrate both sides to get

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