Question

1) \( \int \frac{2}{x^2 - 3x - 18} dx \) 2) \( \int \frac{t + 6}{t^3 - 2t^2} dt \)

          1) \( \int \frac{2}{x^2 - 3x - 18} dx \)
2) \( \int \frac{t + 6}{t^3 - 2t^2} dt \)

1) ∫(2)/(x^2 - 3x - 18) dx
2) ∫(t + 6)/(t^3 - 2t^2) dt

Added by Brian B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert T. L.

01/26/2024

Step 1

Step 1: For the first integral, we have: \int (2)/(x^(2)-3x-18)dx First, we need to factor the denominator: x^2 - 3x - 18 = (x - 6)(x + 3) Now, we can rewrite the integral as: \int (2)/((x - 6)(x + 3))dx Show more…

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int (2)/(x^(2)-3x-18)dx int (t+6)/(t^(3)-2t^(2))dt 1) 2 dx x2-3x-18 2) 9+ dt t3-2t2

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