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a
exxx+4+C
b
exx+2x+4+C
c
exx−2x+4+C
d
2xexx+4+C
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Correct option is A
∫x+2x+42exdx=∫exx2+4x+4(x+4)2dx=∫exx(x+4)(x+4)2+4(x+4)2dx=∫exxx+4+4(x+4)2dx=∫exx(x+4)dx+∫4ex(x+4)2dx=exxx+4−∫4ex(x+4)2dx+∫4ex(x+4)2dx=xex(x+4)+CSimilar Questions
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