Engineering Board Exam Reviewer · Integration by Parts · medium

Evaluate: x2 e3 x,dx

1. ≤ft( x23 - 2 x/9 + 2/27 ) e3 x+C
2. ≤ft( x23 - 2 x/9 - 2/27 ) e3 x+C
3. ≤ft( x23 - 2 x/3 + 2/3 ) e3 x+C
4. ≤ft(x2 - 2 x + 2 ) e3 x9+C

Answer

A. ≤ft( x23 - 2 x/9 + 2/27 ) e3 x+C

Solution

Integrate by parts twice with e3x as the exponential factor (tabular method): x2e3x,dx=e3x≤ft(x2/3-2x/9+2/27 )+C, from the alternating signs +x2· e3x3, -2x· e3x9, +2· e3x27.

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