Question

.∫xe−1+ex−1ex+xedx=∫xe−1+ex−1ex+xedx=

 a)1e.x+c1e.x+c b)e.1x+ce.1x+c c)log (ex + xe) + c d)1e1e. log (ex + xe) + c​

Answer 1

Step-by-step explanation:

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