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a
78(x+2)8−23(x+3)6+C
b
78(x+2)7+23(x+3)6+C
c
(x+2)88+23(x+3)6+C
d
(x+2)88−23(x+3)6+C
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Correct option is D
I=∫(x+3+2)(x+3−2)(x+3)5dx Let x+3=t⇒dx=dtI=∫(t+2)(t−2)t5dtI=∫t7−4t5dtt88−4t66+CSimilar Questions
The value of is equal to
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