Question
Mathematics Question on Integrals of Some Particular Functions
∫(1+xlogx)21+logxdx is equal to
A
1+xlog∣x∣1+C
B
1+log∣x∣1+C
C
1+xlog∣x∣−1+C
D
log1+log∣x∣1+C
Answer
1+xlog∣x∣−1+C
Explanation
Solution
Let I=∫(1+xlogx)21+logxdx Let 1+xlogx=t ⇒(0+x×x1+logx)dx=dt ⇒(1+logx)dx=dt ∴I=∫t21dt=−1t−1+C =(1+xlogx)−1+C