integral pi to 0 log(x+1/x)dx/(1+x^2)
Answers
Answered by
0
Answer:
Let I be the integral given to us.
∴I=∫
0
1
1+x
2
log1+x
dx
Substitute x=tant⟹dx=sec
2
tdt
∴I=∫
0
4
π
sec
2
t
log(1+tant)
sec
2
tdt
∴I=∫
0
4
π
log(1+tant)dt
Now, as ∫
0
a
f(x)dx=∫
0
a
f(a−x)dx, we have
I=∫
0
4
π
log(1+tan(
4
π
−t))dt
I=∫
0
4
π
log(1+
1+tant
1−tant
)dt (using tan(A−B)=
1+tanAtanB
tanA−tanB
)
∴I=∫
0
4
π
log(
1+tant
2
)dt
∴I=∫
0
4
π
[log(2)−log(1+tant)]dt
∴I=∫
0
4
π
log(2)dt−I
∴2I=∫
0
4
π
log(2)dt
∴2I=tlog(2)∣
0
4
π
∴2I=
4
π
log(2)
∴I=
8
π
log(2)
Hence, proved.
Step-by-step explanation:
please do it brianlist answer
Answered by
0
Answer:
the answer π2
Step-by-step explanation:
π√467 is the coking answer and I also love to
Similar questions