Question

="\int\limits {\frac{t-1}{1-t^3} } \, dt" alt="\int\limits {\frac{t-1}{1-t^3} } \, dt" align="absmiddle" class="latex-formula">
Evalue la integral, he intentado sustitucion, por partes, trigonometrica y no he logrado abordarla.

Answer 1

Answer:

-2/√3 atan ((2t + 1)/√3) + C

Explanation:

∫ (t − 1) / (1 − t³) dt

Factor the difference of cubes:

∫ (t − 1) / ((1 − t)(1 + t + t²)) dt

Divide:

∫ -1 / (1 + t + t²) dt

-∫ 1 / (t² + t + 1) dt

Complete the square:

-∫ 1 / (t² + t + ¼ + ¾) dt

-∫ 4 / (4t² + 4t + 1 + 3) dt

-∫ 4 / ((2t + 1)² + 3) dt

If u = 2t + 1, du = 2 dt:

-∫ 2 / (u² + 3) du

Use an integral table, or use trigonometric substitution:

-2 (1/√3) atan (u/√3) + C

-2/√3 atan (u/√3) + C

Substitute back:

-2/√3 atan ((2t + 1)/√3) + C