The evaluation of ∫pxp+2q−1−qxq−1x2p+2q+2xp+q+1dx is
A
−xpxp+q+1+C
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B
xqxp+q+1+C
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C
−xqxp+q+1+C
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D
xpxp+q+1+C
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Solution
The correct option is C−xqxp+q+1+C I=∫pxp+2q−1−qxq−1x2p+2q+2xp+q+1dx=∫pxp−1−qx−q−1(xp+x−q)2dx. Substitute xq+x−q=t⇒(pxp−1−qx−q−1)dx=dt I=∫1t=−1xp+x−q=−xqxp+q+1.