The value of ∫ ( ln ((x-1/x+1))/x2-1) d x is equal to

Question

The value of ∫ ( ln ((x-1/x+1))/x2-1) d x is equal to (A) (1/2) ℓ n2 (x-1/x+1)+C (B) (1/4) ln 2 (x-1/x+1)+C (C) (1/2) ℓ n2 (x+1/x-1)+C (D) (1/4) ℓ n2 (x+1/x-1)+C

Tardigrade Admin · Accepted Answer

I=∫ ( ln ((x-1/x+1))/x2-1) d x put ln ((x-1/x+1))=t ⇒ (2/x2-1) d x=d t ⇒ I=∫ t (d t/2)=(t2/4)+C= log 2((x-1/x+1))+C =(1/4) log 2((x+1/x-1))+C

Q. The value of $\int \frac{l n ( \frac{x - 1}{x + 1} )}{x ^{2} - 1} d x$ is equal to

$\frac{1}{2} ℓ n^{2} \frac{x - 1}{x + 1} + C$

$\frac{1}{4} ln^{2} \frac{x - 1}{x + 1} + C$

$\frac{1}{2} ℓ n^{2} \frac{x + 1}{x - 1} + C$

$\frac{1}{4} ℓ n^{2} \frac{x + 1}{x - 1} + C$

$I = \int \frac{l n ( \frac{x - 1}{x + 1} )}{x ^{2} - 1} d x$
put $ln (\frac{x - 1}{x + 1}) = t \Rightarrow \frac{2}{x ^{2} - 1} d x = d t$
$\Rightarrow I = \int t \frac{d t}{2} = \frac{t ^{2}}{4} + C = lo g^{2} (\frac{x - 1}{x + 1}) + C$
$= \frac{1}{4} lo g^{2} (\frac{x + 1}{x - 1}) + C$

Q. The value of ∫x2−1ln(x+1x−1​)​dx is equal to

21​ℓn2x+1x−1​+C

41​ln2x+1x−1​+C

21​ℓn2x−1x+1​+C

41​ℓn2x−1x+1​+C