Evaluate the ∫e^t dt/cuberoot (1-e^t)
Solution
Substitution x = et :
∫etdt1−et3=∫dx1−x3=∫(1−x)−1/3dx=−32(1−x)2/3+C=−32(1−et)2/3+C\int \frac{e^t dt}{\sqrt[3]{1-e^t}}=\int\frac{dx}{\sqrt[3]{1-x}}=\int(1-x)^{-1/3}dx= -\frac{3}{2}(1-x)^{2/3}+C=-\frac{3}{2}(1-e^t)^{2/3}+C∫31−etetdt=∫31−xdx=∫(1−x)−1/3dx=−23(1−x)2/3+C=−23(1−et)2/3+C
Answer
∫etdt1−et3=−32(1−et)2/3+C\int \frac{e^t dt}{\sqrt[3]{1-e^t}}=-\frac{3}{2}(1-e^t)^{2/3}+C∫31−etetdt=−23(1−et)2/3+C
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment