© examsiri.com
Question : 73 of 160
Marks:
+1,
-0
Solution:
Consider the integral,
I=∫| √x2+1[log(x2+1)−2‌log‌x] |
| x4 |
dx =
∫√1+‌log(1+)‌d‌x Let,
1+=t2 Then,
−2=2tdt So,
I=∫t2‌log‌t2‌d‌t =−2‌∫t2‌log‌t‌d‌t −2[‌log‌t−∫(×)dt] =−t3‌log‌t+t3+c Further simplify the above,
t3[2−3‌log‌t2]+c =(1+)[2−3‌log(1+)]+c
© examsiri.com
Go to Question: