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Find the Partial Fraction Decomposition u=exu = e ^ { x }

Question 134

Multiple Choice

Find the partial fraction decomposition. [Hint: Use the substitution u=exu = e ^ { x } and recall that e2x=(ex) 2]\left. e ^ { 2 x } = \left( e ^ { x } \right) ^ { 2 } \cdot \right]
- 8ex+6e2x5ex6\frac { - 8 e ^ { x } + 6 } { e ^ { 2 x } - 5 e ^ { x } - 6 }


A) 2ex6+6ex+1\frac { - 2 } { e ^ { x } - 6 } + \frac { - 6 } { e ^ { x } + 1 }
B) 6ex6+2ex+1\frac { - 6 } { e ^ { x } - 6 } + \frac { - 2 } { e ^ { x } + 1 }
C) 6e2x6+2e2x+1\frac { - 6 } { e ^ { 2 x } - 6 } + \frac { - 2 } { e ^ { 2 x } + 1 }
D) 2e2x6+6e2x+1\frac { - 2 } { e ^ { 2 x } - 6 } + \frac { - 6 } { e ^ { 2 x } + 1 }

Correct Answer:

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