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The integral ∫((2x)x+(x2)x)log2xdx
Correct Answer: (2x)x+(x2)x+C
Explanation
Let I=∫[(2x)x+(x2)x]ln(2ex)dx
=∫[exlnx−xln2+exln2−xlnx]dx
Let xlnx−xln2=t
(lnx+1−ln2)dx=dt
⇒ln(2ex)dx=dt
∴I=∫(et+e−t)dt
=et−e−t+c
=(2x)x+(x2)x+c
Explanation
Let I=∫[(2x)x+(x2)x]ln(2ex)dx
=∫[exlnx−xln2+exln2−xlnx]dx
Let xlnx−xln2=t
(lnx+1−ln2)dx=dt
⇒ln(2ex)dx=dt
∴I=∫(et+e−t)dt
=et−e−t+c
=(2x)x+(x2)x+c