Question

Evaluate. Assume u > 0 when ln u appears. $\int \frac{e^{1/t^5}}{t^6} dt$ $\circ -e^{1/t^5} + C$ $\circ -\frac{e^{1/t^5}}{5} + C$ $\circ \frac{e^{-1/t^5}}{5} + C$ $\circ -\frac{e^{1/t^5}}{5t^5} + C$

          Evaluate. Assume u > 0 when ln u appears.
$\int \frac{e^{1/t^5}}{t^6} dt$
$\circ -e^{1/t^5} + C$
$\circ -\frac{e^{1/t^5}}{5} + C$
$\circ \frac{e^{-1/t^5}}{5} + C$
$\circ -\frac{e^{1/t^5}}{5t^5} + C$

Evaluate. Assume u > 0 when ln u appears.
∫(e^1/t^5)/(t^6) dt
∘ -e^1/t^5 + C
∘ -(e^1/t^5)/(5) + C
∘(e^-1/t^5)/(5) + C
∘ -(e^1/t^5)/(5t^5) + C

Added by Jean B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Muhammed Suhaid

10/21/2020

Step 1

Step 1: Rewrite the integral as ∫ t^(-5) + 6 dt Show more…

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Evaluate. Assume u > 0 when ln u appears. ∫ 1/t^5 + 6 dt = -e^(-1/t^5) + C = e^(1/t^5) + C = 5e^(-1/t^5) + C = 5e^(1/t^5) + C

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Question

Evaluate. Assume u > 0 when ln u appears. $\int \frac{e^{1/t^5}}{t^6} dt$ $\circ -e^{1/t^5} + C$ $\circ -\frac{e^{1/t^5}}{5} + C$ $\circ \frac{e^{-1/t^5}}{5} + C$ $\circ -\frac{e^{1/t^5}}{5t^5} + C$

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