integral value of 2xdx /(x+1)(x-1))
Answers
Answered by
0
Answer:
log |x^2 - 1| + C
Step-by-step explanation:
Solving the denominator:
Let, (x + 1)(x - 1) = t
x^2 - 1 = t , differentiate w.r.t x:
=> d(x^2 - 1)/dx = dt/dx
=> 2x dx = dt
Therefore,
=> 2x dx / (x + 1)(x - 1)
=> dt/t
=> log |t| + C
=> log |x^2 - 1| + C
Which can also be written as ln |x^2 - 1|
Answered by
0
Answer:
Let I=∫011−x1+x−−−√dx
=∫011−x1−x2√dx
=∫0111−x2√dx−∫01x1+x2√dx
=[sin−1x]10−∫01x1−x2√dx
Put t2=1−x2⇒2tdt=−2xdx
⇒tdt=−xdx
∴I=(sin−11−sin−10)+∫10ttdt
=π2+[t]01=π2−1
Step-by-step explanation:
Similar questions
=∫011−x1−x2√dx
=∫0111−x2√dx−∫01x1+x2√dx
=[sin−1x]10−∫01x1−x2√dx
Put t2=1−x2⇒2tdt=−2xdx
⇒tdt=−xdx
∴I=(sin−11−sin−10)+∫10ttdt
=π2+[t]01=π2−1