Please solve the following question

Asked by Balbir | 7th Oct, 2017, 11:20: PM

Expert Answer:

begin mathsize 16px style straight I equals integral straight a to the power of straight x space open square brackets logx space plus space loga space log open parentheses straight x to the power of straight x over straight e to the power of straight x close parentheses close square brackets dx straight I equals integral straight a to the power of straight x space open square brackets logx space plus space loga space open parentheses logx to the power of straight x space minus space loge to the power of straight x close parentheses close square brackets dx straight I equals integral straight a to the power of straight x space open square brackets logx space plus space loga space open parentheses xlogx space minus space straight x close parentheses close square brackets dx straight I equals integral open square brackets loga space open parentheses xlogx space minus space straight x close parentheses space plus space straight a to the power of straight x space logx close square brackets space dx For space the space integral space straight a to the power of straight x space logx space integrate space by space parts space where straight u space equals space straight a to the power of straight x space and space straight v space equals space log space straight x For space the space integral space loga space open parentheses xlogx space minus space straight x close parentheses space we space get space straight a to the power of straight x space open parentheses xlogx space minus straight x close parentheses end style

Answered by Sneha shidid | 18th Dec, 2017, 10:53: AM

Related Videos

Integration by partial fractions when factors in denominator are linear bu...

Integration by partial fractions when factors in denominator are linear bu...

Integration by partial fractions when factors in denominator are linear bu...

All Questions Ask Doubt