Question
If ∫(x−1x+1)dx√x3+x2+x=2tan−1√f(x)+C, find f(x).
Answer: Option A
:
A
I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
Was this answer helpful ?
:
A
I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
Was this answer helpful ?
More Questions on This Topic :
Question 2. ∫(2+sec x)sec x(1+2sec x)2dx....
Question 3. ∫(1+√tanx)(1+tan2x)2tanxdx equal to ....
Question 5. ∫√1+3√x3√x2dx is equal to....
Question 6. ∫x2−2x3√x2−1dx is equal to....
Question 9. ∫sin3x(cos4x+3cos2x+1)tan−1(secx+cosx)dx=....
Submit Solution