Answer:
Explanation:
I am attempting to solve the following integral, in which I am not sure, how to proceed:
∫1u2−2du
I have tried the following, but I am not sure if it is correct:
y=u2−2
dy=2u∗du
du=dy2u
Therefore:
∫1u2−1du=∫1y∗dy2u=∫1y∗12udy=∫1y2udy
Solving for u:
u2=y+2
u=y+2−−−−√
Substituting u for y+2−−−−√:
∫1y2udy=∫1y2(y+2√)dy
I am not sure, where to go from here. The only thing I can remember is to use the logarithm rule where:
∫1y2(y+2√)dy=ln|y2(y+2−−−−√)|+C
2
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Answers & Comments
Answer:
Explanation:
I am attempting to solve the following integral, in which I am not sure, how to proceed:
∫1u2−2du
I have tried the following, but I am not sure if it is correct:
y=u2−2
dy=2u∗du
du=dy2u
Therefore:
∫1u2−1du=∫1y∗dy2u=∫1y∗12udy=∫1y2udy
Solving for u:
u2=y+2
u=y+2−−−−√
Substituting u for y+2−−−−√:
∫1y2udy=∫1y2(y+2√)dy
I am not sure, where to go from here. The only thing I can remember is to use the logarithm rule where:
∫1y2(y+2√)dy=ln|y2(y+2−−−−√)|+C
Verified answer
2
I am attempting to solve the following integral, in which I am not sure, how to proceed:
∫1u2−2du
I have tried the following, but I am not sure if it is correct:
y=u2−2
dy=2u∗du
du=dy2u
Therefore:
∫1u2−1du=∫1y∗dy2u=∫1y∗12udy=∫1y2udy
Solving for u:
u2=y+2
u=y+2−−−−√
Substituting u for y+2−−−−√:
∫1y2udy=∫1y2(y+2√)dy
I am not sure, where to go from here. The only thing I can remember is to use the logarithm rule where:
∫1y2(y+2√)dy=ln|y2(y+2−−−−√)|+C