Question
Mathematics Question on integral
Evaluate : ∫(exloga+ealogx+ealoga)dx
A
logaax+a+1xa+1+aax+c
B
logaax+a−1xa+1+axa+c
C
logaax+a+1xa+1+axa+c
D
logaax−a+1xa+1+aax+c
Answer
logaax+a+1xa+1+aax+c
Explanation
Solution
We have,I=∫(exloga+ealogx+ealoga)dx Then,I=∫(elogax+elogxa+elogaa)dx ⇒I=∫(ax+xa+aa)dx[∵elogλ=λ] ⇒∫axdx+∫xadx+∫aadx ⇒I=logaax+a+1xa+1+aax+C